Subject: Calculus

# One Sided Limits

## Calculus.One-sidedLimit History

Hide minor edits - Show changes to output

June 26, 2011
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Changed line 73 from:

!!Example #1

to:

!!!Example #1

Added lines 104-105:

'''NEXT TOPIC:'''[[Limit of a sequence]]

December 31, 2010
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December 31, 2010
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November 01, 2010
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Changed lines 44-46 from:

For every {$ \~~ (~~epsilon~~ ~~>0) $} (no matter how small) there will exist a {$ (\delta>0) $} such that {$ |f(x)-L|< \epsilon$} whenever {$ [0< x-a <\delta] $}.

to:

For every {$ (\epsilon>0) $} (no matter how small) there will exist a {$ (\delta>0) $} such that {$ |f(x)-L|< \epsilon$} whenever {$ [0< x-a <\delta] $}.

Changed line 66 from:

For every {$ \~~ (~~epsilon >0) $} (no matter how small) there will exist a {$ (\delta>0) $} such that {$ |f(x)-L|< \epsilon$} whenever {$ [0< x-a <\delta] $}.

to:

For every {$ (\epsilon >0) $} (no matter how small) there will exist a {$ (\delta>0) $} such that {$ |f(x)-L|< \epsilon$} whenever {$ [0< x-a <\delta] $}.

November 01, 2010
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Changed lines 2-3 from:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we weren't sure of the direction approaching {$a$} (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\

to:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we weren't sure of the direction approaching {$a$} (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\\

Changed lines 7-10 from:

One-sided limits are important when there is a discontinuity of the function on either side of {$a$}. When that situation occurs, a two-sided limit will not work. It may be true on one side but it can't be on the other side. An example would do a better job of explaining this: consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which come in two flavors. \\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which come in two flavors. \\

to:

One-sided limits are important when there is a discontinuity of the function on either side of {$a$}. When that situation occurs, a two-sided limit will not work. It may be true on one side but it can't be on the other side. An example would do a better job of explaining this: consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which come in two flavors. \\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which come in two flavors. \\\

November 01, 2010
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Changed line 2 from:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we weren't ~~sureof~~ the direction approaching {$a$} (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\

to:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we weren't sure of the direction approaching {$a$} (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\

November 01, 2010
by -

Changed lines 2-3 from:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we ~~had~~ ~~no~~ ~~predilection~~ ~~for~~ ~~which~~ ~~direction approached ~~{$a$}~~ from~~ (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\

to:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we weren't sureof the direction approaching {$a$} (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\

Changed lines 6-11 from:

One-sided limits are important when there is a discontinuity of the function on either side of {$a$}. When that occurs a two-sided limit will not work. It may be true on one side but it can't be on the other side. An example would do a better job of explaining this: consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which~~com~~ in two flavors. \\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which

to:

One-sided limits are important when there is a discontinuity of the function on either side of {$a$}. When that situation occurs, a two-sided limit will not work. It may be true on one side but it can't be on the other side. An example would do a better job of explaining this: consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which come in two flavors. \\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which come in two flavors. \\

Changed lines 30-31 from:

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}. Knowing this, we can now define a right-handed limit

to:

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}. Knowing this, we can now define a right-handed limit.

Changed line 77 from:

''Solution:'' Notice that if we ~~follow ~~directly theorems discussed before by substituting ~~directly ~~1 to the function, it gives zero at the denominator. To avoid these, let us rationalize the denominator. Note that rationalizing denominator is not a standard option in solving one-sided limits. The use of ~~such~~ is on a case to case basis. Well anyway, let's proceed.\\

to:

''Solution:'' Notice that if we directly follow theorems discussed before by substituting 1 to the function, it gives zero at the denominator. To avoid these, let us rationalize the denominator. Note that rationalizing the denominator is not a standard option in solving one-sided limits. The use of this method is on a case to case basis. Well anyway, let's proceed.\\

Changed line 100 from:

In a broader sense, the solutions and techniques ~~employed~~ ~~in~~ ~~solving~~ two-sided limits are still applicable for one-sided limits, as well as the [[Limit of a function | theorems]] involved.

to:

In a broader sense, the solutions and techniques we use to solve two-sided limits are still applicable for one-sided limits, as well as the [[Limit of a function | theorems]] involved.

October 27, 2010
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October 27, 2010
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Changed line 82 from:

|| || = ||'+{$ ~~{~~\frac{x-1}{\sqrt{ (x+1)(x-1)~~ ~~}}~~}~~ ~~x~~ ~~{\frac{\sqrt{~~ ~~(x-1)~~ ~~(x+1) ~~}~~}~~{\sqrt{ (x+1) (x-1) }}} $}+' ||

to:

|| || = ||'+{$ \frac{x-1}{\sqrt{ (x+1)(x-1)}} * $} {$ {\sqrt{ (x-1) (x+1)}\over {\sqrt{ (x+1) (x-1) }}} $}+' ||

Changed line 93 from:

|| '+{$\lim_{x \to 1^{+}} f(x)$}+'|| = ||'+{$\lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}{ (x+1) }$}+' ||

to:

|| '+{$\lim_{x \to 1^{+}} f(x)$}+'|| = ||'+{$\lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}\over{ (x+1) }$}+' ||

October 26, 2010
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Changed line 82 from:

|| || = ||'+{$ ~~\frac~~{x-1}{\sqrt{ (x+1)(x-1) }} x ~~\frac~~{\~~sqrt~~{ (x-1) (x+1) }}{\sqrt{ (x+1) (x-1) }} $}+' ||

to:

|| || = ||'+{$ {\frac{x-1}{\sqrt{ (x+1)(x-1) }}} x {\frac{\sqrt{ (x-1) (x+1) }}{\sqrt{ (x+1) (x-1) }}} $}+' ||

October 26, 2010
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Changed lines 2-15 from:

In the ~~previous~~ ~~lessons,~~ ~~we~~ ~~are~~ ~~concerned~~ ~~with~~ ~~the~~ ~~values~~ ~~of~~ ~~{$~~ ~~x~~ ~~\neq~~ ~~a~~ ~~$}~~ ~~when~~ ~~considering~~ ~~the~~ ~~limit~~ ~~of {~~$~~f$~~} ~~as~~ ~~it~~ ~~approaches~~ ~~{$a$}~~ ~~in~~ ~~an~~ ~~open~~ ~~interval ~~{$~~I~~$} ~~that~~ ~~are~~ ~~close~~ ~~and yet, contains ~~{$a$}~~.~~ ~~We~~ ~~did~~ ~~work~~ ~~out~~ ~~on~~ ~~those~~ ~~limits~~ ~~without~~ ~~being~~ ~~particular~~ ~~to~~ ~~which~~ ~~direction~~ ~~the~~ ~~approach~~ ~~is~~ ~~coming.~~ ~~We~~ ~~just~~ ~~consider~~ ~~both~~ ~~values of ~~{$~~x~~$} ~~coming ~~from ~~the~~ ~~left~~ ~~and~~ ~~right~~ of ~~{$a$}.~~ ~~This~~ ~~kind~~ ~~of~~ ~~mechanism~~ ~~is~~ ~~attributed~~ ~~to~~ ~~two-sided~~ ~~limits.~~ ~~\\~~

[[<<]]

!!Why ~~is~~ ~~there~~ ~~a~~ ~~single-sided~~ ~~limits?~~

%rframe ~~width=300px~~ ~~border='0px~~ ~~solid~~ ~~black'~~ ~~padding=2px%Attach:one_sided_limits~~.~~jpg~~ ~~|~~ ~~'''Figure~~ ~~1:~~ ~~Plot~~ of ~~the~~ ~~function~~ ~~{$f(x)$}~~. ~~[[http://www.mathematics2.com/Calculus/One-sidedLimit?action=download&upname=one_sided_limits.jpg~~ ~~|~~ (~~Source)]]'''~~

One ~~sided~~ ~~limits~~ ~~are~~ ~~considered~~ ~~and~~ ~~is~~ ~~important~~ ~~because~~ ~~when~~ ~~there~~ ~~is~~ ~~a~~ ~~discontinuity~~ ~~of~~ ~~the~~ ~~function~~ ~~on~~ ~~either~~ ~~side~~ ~~of~~ ~~{$~~a~~$},~~ ~~then~~ ~~the~~ ~~two-sided~~ ~~limits~~ ~~is~~ ~~not~~ ~~true~~ ~~at~~ ~~all.~~ ~~It~~ ~~may~~ ~~be~~ ~~true~~ ~~on~~ ~~one~~ ~~side but it can't be on the other side~~. ~~As~~ ~~an illustration, consider the function {$f~~(~~x~~)~~=\sqrt{1+x}$}~~ ~~which~~ ~~is~~ ~~plotted~~ ~~in~~ ~~Figure~~ ~~1.\\~~

Note ~~that~~ ~~the~~ ~~value ~~of the function ~~is~~ ~~imaginary~~ ~~at~~ ~~points~~ {$~~x<-1~~$} ~~as~~ ~~seen~~ ~~in~~ ~~the~~ ~~figure.~~ ~~This~~ ~~corresponds~~ ~~to~~ ~~the~~ ~~left~~ ~~side~~ ~~of~~ ~~the~~ ~~the~~ ~~limit~~ ~~supposing~~ ~~that~~ ~~{$a=-1$}.~~ ~~Hence,~~ ~~{$f(x)$}~~ ~~is~~ ~~not~~ ~~defined~~ ~~on~~ ~~the~~ ~~left~~ ~~side~~ ~~of~~ ~~{$~~a~~$},~~ ~~thus,~~ ~~the~~ ~~two~~ ~~sided~~ ~~limit~~ ~~does~~ ~~not~~ ~~makes~~ ~~sense~~ ~~in~~ ~~this~~ ~~type~~ ~~of~~ ~~configuration.~~ ~~However,~~ ~~if~~ ~~we~~ ~~consider~~ the ~~values~~ ~~of~~ {$x$} ~~that~~ ~~are~~ ~~greater~~ ~~than~~ ~~-1,~~ ~~then~~ ~~we~~ ~~find~~ ~~that ~~the ~~values~~ ~~of~~ {$~~f~~$} ~~are~~ ~~defined!~~ ~~See!~~ ~~So~~ ~~this~~ ~~means~~ ~~that~~ ~~on~~ ~~one-~~side~~,~~ ~~the~~ ~~limit ~~{$~~\lim_{x~~ ~~\to~~ ~~-1^{+}}\sqrt{1+x}$}~~ ~~exists.~~ ~~Thus,~~ ~~it~~ ~~is~~ ~~important~~ ~~to~~ ~~take~~ ~~time~~ ~~studying~~ ~~not~~ ~~just~~ ~~the~~ ~~two-sided~~ ~~limits~~ ~~but~~ ~~also~~ ~~one-sided~~ ~~limits~~ ~~as~~ ~~well~~ ~~which~~ ~~comes~~ ~~in~~ ~~two~~ ~~types.~~ ~~\\~~

[[<<]]

!!Classifications of One-~~Sided~~ ~~Limits~~

[[<<]]

!!Why

%rframe

One

Note

[[<<]]

!!Classifications of One

to:

In the [[limit of a function | previous lesson]], we were concerned with the values of {$ x \neq a $} when considering the limit of the function {$f$} as it approaches {$a$} in the open interval {$I$} that contains {$a$} (what a mouthful). For those limits, we had no predilection for which direction approached {$a$} from (be it the left side of the limit or the right side). We just considered both values of {$x$} coming from the left and right of {$a$}. This is the standard definition of a two-sided limit. But as you can guess (from the title of this page I hope), there are also one-sided limits that will be addressed here.\\

!!You lost me. Why would there be a single-sided limit?

%rframe width=300px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | ''''-Figure 1: Plot of the function {$f(x)$}. [[http://www.mathematics2.com/Calculus/One-sidedLimit?action=download&upname=one_sided_limits.jpg | (Source)]]-''''

One-sided limits are important when there is a discontinuity of the function on either side of {$a$}. When that occurs a two-sided limit will not work. It may be true on one side but it can't be on the other side. An example would do a better job of explaining this: consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which com in two flavors. \\

!!You lost me. Why would there be a single-sided limit?

%rframe width=300px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | ''''-Figure 1: Plot of the function {$f(x)$}. [[http://www.mathematics2.com/Calculus/One-sidedLimit?action=download&upname=one_sided_limits.jpg | (Source)]]-''''

One-sided limits are important when there is a discontinuity of the function on either side of {$a$}. When that occurs a two-sided limit will not work. It may be true on one side but it can't be on the other side. An example would do a better job of explaining this: consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary when {$x<-1$}. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}. So, a two sided limit would not be applicable to this type of problem. However, if we consider the values of {$x$} that are greater than -1, we find that the values of {$f$} are defined! See! So this means that on one side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which com in two flavors. \\

Changed line 13 from:

(:cellnr bgcolor=#d4d7ba colspan=14 align=center:) '+'''Two Types of One-Sided Limits'''+'

to:

(:cellnr bgcolor=#d4d7ba colspan=14 align=center:) '+'''The Two Types of One-Sided Limits'''+'

Changed line 16 from:

*[[#t2 | Right-Hand Limit: Definition]]

to:

**[[#t2 | Right-Hand Limit: Definition]]

Changed lines 18-19 from:

*[[#t4 | Left-Hand Limit: Definition]]

to:

**[[#t4 | Left-Hand Limit: Definition]]

* [[#ex1 | An example of a one-sided limit]]

* [[#ex1 | An example of a one-sided limit]]

Changed lines 24-25 from:

As mentioned, there are two types of one-sided limits. The first type is when the value of {$x$} approaches from the right of {$a$}~~.~~ ~~This~~ ~~is~~ ~~so~~ ~~because~~ we ~~are~~ ~~considering~~ ~~the~~ ~~values~~ of ~~{$x$}~~ ~~to~~ ~~the~~ ~~right~~ ~~of~~ {$~~a~~$}~~.~~ ~~This~~ ~~only~~ ~~happens~~ ~~when~~ {$~~f~~$} ~~is properly defined on that side.~~ In limit notation, this goes like,

to:

As mentioned, there are two types of one-sided limits. '''The first type''' is when the value of {$x$} approaches from the right of {$a$} (similar to the example we showed you above). This type of limit would only consider the values of {$x$} to the right of {$a$}. In limit notation, this goes like,

Changed lines 27-31 from:

\\~~\~~

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}.~~Having~~ ~~known ~~this~~ one~~, ~~let~~ ~~us~~ ~~know~~ define right-handed limit~~ formally.~~

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}.

to:

\\

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}. Knowing this, we can now define a right-handed limit

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}. Knowing this, we can now define a right-handed limit

Changed lines 33-35 from:

!!Right-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is{$L$}, notated as

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is

to:

!!!Right-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is equal to {$L$}", notated as:

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is equal to {$L$}", notated as:

Changed lines 41-43 from:

to:

For every {$ \ (epsilon >0) $} (no matter how small) there will exist a {$ (\delta>0) $} such that {$ |f(x)-L|< \epsilon$} whenever {$ [0< x-a <\delta] $}.

Changed lines 46-47 from:

Likewise, ~~when~~ ~~"L~~ ~~is ~~the limit of {$f(x)$} as {$x$} approaches a from the left, ~~then~~ we ~~may ~~write ~~it~~ as

to:

Likewise, "L can be the limit of {$f(x)$} as {$x$} approaches a from the left", and we'll write this as:

Changed lines 49-53 from:

\\~~\~~

The symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define left-sided limits formally~~.~~

The symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define left-sided limits formally

to:

\\

The symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define left-sided limits formally:

The symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define left-sided limits formally:

Changed lines 55-58 from:

!!Left-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (c,a). Then "the limit of {$f(x)$} as {$x$} approaches {$a$} from the left is{$L$}, notated as

Let {$f$} be a function defined at every point in some open interval (c,a). Then "the limit of {$f(x)$} as {$x$} approaches {$a$} from the left is

to:

!!!Left-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (c,a). Then "the limit of {$f(x)$} as {$x$} approaches {$a$} from the left is equal to {$L$}", notated as

Let {$f$} be a function defined at every point in some open interval (c,a). Then "the limit of {$f(x)$} as {$x$} approaches {$a$} from the left is equal to {$L$}", notated as

Changed lines 63-67 from:

to:

For every {$ \ (epsilon >0) $} (no matter how small) there will exist a {$ (\delta>0) $} such that {$ |f(x)-L|< \epsilon$} whenever {$ [0< x-a <\delta] $}.

Changed lines 67-69 from:

!!~~!~~Example #1

Evaluate the following~~limits~~:

Evaluate the following

to:

\\\

[[#ex1]]

!!Example #1

Evaluate the following limit:

[[#ex1]]

!!Example #1

Evaluate the following limit:

Changed line 82 from:

|| || = ||'+{$ \frac{x-1}{\sqrt{ (x+1)(x-1) }} ~~\cdot~~\frac{\sqrt{ (x-1) (x+1) }}{\sqrt{ (x+1) (x-1) }} $}+' ||

to:

|| || = ||'+{$ \frac{x-1}{\sqrt{ (x+1)(x-1) }} x \frac{\sqrt{ (x-1) (x+1) }}{\sqrt{ (x+1) (x-1) }} $}+' ||

Changed line 100 from:

In a broader sense, the solutions and techniques employed in solving two-sided limits are still applicable for one-sided limits, as well as the ~~theorems~~ involved.

to:

In a broader sense, the solutions and techniques employed in solving two-sided limits are still applicable for one-sided limits, as well as the [[Limit of a function | theorems]] involved.

October 26, 2010
by -

Changed line 2 from:

In the previous lessons, we are concerned with the values of {$x \neq a$} when considering the limit of {$f$} as it approaches {$a$} in an open interval {$I$} that are close and yet, contains {$a$}. We did work out on those limits without being particular to which direction the approach is coming. We just consider both values of {$x$} coming from the left and right of {$a$}. This kind of mechanism is attributed to two-sided limits. \\

to:

In the previous lessons, we are concerned with the values of {$ x \neq a $} when considering the limit of {$f$} as it approaches {$a$} in an open interval {$I$} that are close and yet, contains {$a$}. We did work out on those limits without being particular to which direction the approach is coming. We just consider both values of {$x$} coming from the left and right of {$a$}. This kind of mechanism is attributed to two-sided limits. \\

October 26, 2010
by -

Changed line 1 from:

(:title~~:~~ One Sided Limits:)

to:

(:title One Sided Limits:)

October 26, 2010
by -

Changed line 73 from:

to:

!!!Example #1

Changed lines 76-108 from:

\end{example}

\indent

Notice

\newline{}

\indent

First,

\begin{eqnarray}

We

\begin{eqnarray}

\end

Thus,

\end{document}

to:

'+{$$ \lim_{x \to 1^{+}} \frac{x-1}{\sqrt{x^{2}-1}}$$}+'

(:toggle id=box1 show='Show Answers with Solutions' init=hide button=1:)

>>id=box1 border='1px solid #999' padding=5px bgcolor=#edf<<

''Solution:'' Notice that if we follow directly theorems discussed before by substituting directly 1 to the function, it gives zero at the denominator. To avoid these, let us rationalize the denominator. Note that rationalizing denominator is not a standard option in solving one-sided limits. The use of such is on a case to case basis. Well anyway, let's proceed.\\

First, let us rationalize the function

|| border=0 align=center width=80%

|| '+{$f(x) $}+'|| = ||'+{$ \frac{x-1}{\sqrt{x^{2}-1}} $}+' ||

|| || = ||'+{$ \frac{x-1}{\sqrt{ (x+1)(x-1) }} \cdot\frac{\sqrt{ (x-1) (x+1) }}{\sqrt{ (x+1) (x-1) }} $}+' ||

|| || = ||'+{$ \frac{(x-1)\sqrt{x^{2}-1}}{\sqrt{ (x+1)^{2}(x-1)^{2}}} $}+' ||

|| || = ||'+{$ \frac{(x-1)\sqrt{x^{2}-1}}{ (x+1)(x-1) } $}+' ||

|| || = ||'+{$ \frac{\sqrt{x^{2}-1}}{ (x+1) } $}+' ||

We have now succesfully rationalized the denominator. Observe that it is not zero at the denominator when {$x$} is replaced by {$a$}.

Thus, we now proceed accordingly.

|| border=0 align=center width=80%

|| '+{$\lim_{x \to 1^{+}} f(x)$}+'|| = ||'+{$\lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}{ (x+1) }$}+' ||

|| || = ||'+{$\frac{\sqrt{1^{2}-1}}{1+1}$}+' ||

|| || = ||'+{$\frac{0}{2}$}+' ||

|| || = ||'+0+' ||

>><<

In a broader sense, the solutions and techniques employed in solving two-sided limits are still applicable for one-sided limits, as well as the theorems involved.

(:toggle id=box1 show='Show Answers with Solutions' init=hide button=1:)

>>id=box1 border='1px solid #999' padding=5px bgcolor=#edf<<

''Solution:'' Notice that if we follow directly theorems discussed before by substituting directly 1 to the function, it gives zero at the denominator. To avoid these, let us rationalize the denominator. Note that rationalizing denominator is not a standard option in solving one-sided limits. The use of such is on a case to case basis. Well anyway, let's proceed.\\

First, let us rationalize the function

|| border=0 align=center width=80%

|| '+{$f(x) $}+'|| = ||'+{$ \frac{x-1}{\sqrt{x^{2}-1}} $}+' ||

|| || = ||'+{$ \frac{x-1}{\sqrt{ (x+1)(x-1) }} \cdot\frac{\sqrt{ (x-1) (x+1) }}{\sqrt{ (x+1) (x-1) }} $}+' ||

|| || = ||'+{$ \frac{(x-1)\sqrt{x^{2}-1}}{\sqrt{ (x+1)^{2}(x-1)^{2}}} $}+' ||

|| || = ||'+{$ \frac{(x-1)\sqrt{x^{2}-1}}{ (x+1)(x-1) } $}+' ||

|| || = ||'+{$ \frac{\sqrt{x^{2}-1}}{ (x+1) } $}+' ||

We have now succesfully rationalized the denominator. Observe that it is not zero at the denominator when {$x$} is replaced by {$a$}.

Thus, we now proceed accordingly.

|| border=0 align=center width=80%

|| '+{$\lim_{x \to 1^{+}} f(x)$}+'|| = ||'+{$\lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}{ (x+1) }$}+' ||

|| || = ||'+{$\frac{\sqrt{1^{2}-1}}{1+1}$}+' ||

|| || = ||'+{$\frac{0}{2}$}+' ||

|| || = ||'+0+' ||

>><<

In a broader sense, the solutions and techniques employed in solving two-sided limits are still applicable for one-sided limits, as well as the theorems involved.

October 26, 2010
by -

Changed line 5 from:

%rframe width=~~200px~~ border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | '''Figure 1: Plot of the function {$f(x)$}.'''

to:

%rframe width=300px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | '''Figure 1: Plot of the function {$f(x)$}. [[http://www.mathematics2.com/Calculus/One-sidedLimit?action=download&upname=one_sided_limits.jpg | (Source)]]'''

October 26, 2010
by -

Changed lines 19-21 from:

*[[#t2 | ~~Left~~-~~Handed~~ Limit]]

to:

*[[#t2 | Right-Hand Limit: Definition]]

*[[#t3 | Left-Handed Limit]]

*[[#t4 | Left-Hand Limit: Definition]]

*[[#t3 | Left-Handed Limit]]

*[[#t4 | Left-Hand Limit: Definition]]

Added lines 24-25:

[[#t1]]

!! Right-Handed Limit

!! Right-Handed Limit

Changed lines 28-29 from:

to:

%cframe align=center%'+{$ \lim_{x \to a^{+}} f(x)=L $}+'

\\\

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}. Having known this one, let us know define right-handed limit formally.

[[#t2]]

!!Right-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is {$L$}, notated as

\\\

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$}. Having known this one, let us know define right-handed limit formally.

[[#t2]]

!!Right-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is {$L$}, notated as

Changed lines 43-45 from:

to:

if for every {$\epsilon >0$}, no matter how small, there exists a {$\delta>0$} such that {$|f(x)-L|< \epsilon$} whenever {$0<x-a<\delta$}.

[[#t3]]

!!Left-Handed Limit

Likewise, when "L is the limit of {$f(x)$} as {$x$} approaches a from the left, then we may write it as

%cframe align=center%'+{$ \lim_{x \to a^{-}} f(x)=L $}+'

\\\

The symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define left-sided limits formally.

[[#t4]]

!!Left-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (c,a). Then "the limit of {$f(x)$} as {$x$} approaches {$a$} from the left is {$L$}, notated as

[[#t3]]

!!Left-Handed Limit

Likewise, when "L is the limit of {$f(x)$} as {$x$} approaches a from the left, then we may write it as

%cframe align=center%'+{$ \lim_{x \to a^{-}} f(x)=L $}+'

\\\

The symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define left-sided limits formally.

[[#t4]]

!!Left-Hand Side Limit: Definition

Let {$f$} be a function defined at every point in some open interval (c,a). Then "the limit of {$f(x)$} as {$x$} approaches {$a$} from the left is {$L$}, notated as

Changed line 63 from:

'+{$$ \lim_{x \to a^{~~-~~}} f(x)=L $$}+'

to:

'+{$$ \lim_{x \to a^{+}} f(x)=L $$}+'

Changed lines 67-89 from:

[[#t1]]

!!Right-Hand

Let

\begin{equation}

\lim_{x \to a^{+}} f(x)=L

\end{equation}

if for every $\epsilon >0$, no matter how small, there exists a $\delta>0$ such that $|f(x)-L|< \epsilon$ whenever $0<x-a<\delta$.

\end{definition}

\begin{definition}[Left-Hand Side Limit]

Let $f$ be a function defined at every point in some open interval (c,a). Then \textbf{the limit of f(x) as x approaches a from the left is L}, notated as

\begin{equation}

\lim_{x \to a^{-}} f(x)=L

\end{equation}

if for every $\epsilon >0$, no matter how small, there exists a $\delta>0$ such that $|f(x)-L|< \epsilon$ whenever $0<x-a<\delta$.

\end{definition}

\indent

As an important remark, the theorems discussed in the previous article still holds true for one-sided

to:

if for every {$\epsilon >0$}, no matter how small, there exists a {$\delta>0$} such that {$|f(x)-L|< \epsilon$} whenever {$0<x-a<\delta$}.

%lframe text-align=center%As an important remark, the theorems discussed in the [[ LimitOfAFunction | previous article ]] still holds true for one-sided limits where "{$x-a$}" is replaced by "{$x-a^{+}$}" or "{$x-a^{-}$}".

\begin{example}

Evaluate the following limits:

%lframe text-align=center%As an important remark, the theorems discussed in the [[ LimitOfAFunction | previous article ]] still holds true for one-sided limits where "{$x-a$}" is replaced by "{$x-a^{+}$}" or "{$x-a^{-}$}".

\begin{example}

Evaluate the following limits:

October 26, 2010
by -

Changed lines 3-13 from:

Note that the value of the function is imaginary at points {$x<-1$} as seen in the figure. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}, thus, the two sided limit does not

Obviously,

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

to:

!!Why is there a single-sided limits?

%rframe width=200px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | '''Figure 1: Plot of the function {$f(x)$}.'''

One sided limits are considered and is important because when there is a discontinuity of the function on either side of {$a$}, then the two-sided limits is not true at all. It may be true on one side but it can't be on the other side. As an illustration, consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary at points {$x<-1$} as seen in the figure. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}, thus, the two sided limit does not makes sense in this type of configuration. However, if we consider the values of {$x$} that are greater than -1, then we find that the values of {$f$} are defined! See! So this means that on one-side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which comes in two types. \\

[[<<]]

!!Classifications of One-Sided Limits

(:table border=1 cellpadding=5 cellspacing=0:)

(:cellnr bgcolor=#d4d7ba colspan=14 align=center:) '+'''Two Types of One-Sided Limits'''+'

%rframe width=200px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | '''Figure 1: Plot of the function {$f(x)$}.'''

One sided limits are considered and is important because when there is a discontinuity of the function on either side of {$a$}, then the two-sided limits is not true at all. It may be true on one side but it can't be on the other side. As an illustration, consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary at points {$x<-1$} as seen in the figure. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}, thus, the two sided limit does not makes sense in this type of configuration. However, if we consider the values of {$x$} that are greater than -1, then we find that the values of {$f$} are defined! See! So this means that on one-side, the limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} exists. Thus, it is important to take time studying not just the two-sided limits but also one-sided limits as well which comes in two types. \\

[[<<]]

!!Classifications of One-Sided Limits

(:table border=1 cellpadding=5 cellspacing=0:)

(:cellnr bgcolor=#d4d7ba colspan=14 align=center:) '+'''Two Types of One-Sided Limits'''+'

Changed lines 18-19 from:

to:

*[[#t1 | Right-Handed Limit]]

*[[#t2 | Left-Handed Limit]]

*[[#t2 | Left-Handed Limit]]

Changed lines 22-24 from:

which

to:

As mentioned, there are two types of one-sided limits. The first type is when the value of {$x$} approaches from the right of {$a$}. This is so because we are considering the values of {$x$} to the right of {$a$}. This only happens when {$f$} is properly defined on that side. In limit notation, this goes like,

Changed line 28 from:

'+{$$ \lim_{x \to a^{~~-~~}} f(x)=L $$}+'

to:

'+{$$ \lim_{x \to a^{+}} f(x)=L $$}+'

Added lines 32-39:

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". Likewise, when "L is the limit of {$f(x)$} as {$x$} approaches a from the left, then we may write it as

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

'+{$$ \lim_{x \to a^{-}} f(x)=L $$}+'

(:tableend:)

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

'+{$$ \lim_{x \to a^{-}} f(x)=L $$}+'

(:tableend:)

Changed lines 42-43 from:

to:

[[#t1]]

!!Right-Hand Side Limit

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is {$L$}, notated as

!!Right-Hand Side Limit

Let {$f$} be a function defined at every point in some open interval (a,b). Then "the limit of {$f(x)$} as {$x$} approaches a from the right is {$L$}, notated as

October 26, 2010
by -

Added lines 1-85:

In the previous lessons, we are concerned with the values of {$x \neq a$} when considering the limit of {$f$} as it approaches {$a$} in an open interval {$I$} that are close and yet, contains {$a$}. We did work out on those limits without being particular to which direction the approach is coming. We just consider both values of {$x$} coming from the left and right of {$a$}. This kind of mechanism is attributed to two-sided limits. \\

%rframe width=250px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | '''Figure 1: Plot of the function {$f(x)=\sqrt{1+x}$}.'''

On the other hand, one sided limits are also considered here. This kind of limits are important because when there is a discontinuity of the function on either side of $a$, then the two-sided limits is not true at all. It may be true on one side but it can't be on the other side. As an illustration, consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary at points {$x<-1$} as seen in the figure. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}, thus, the two sided limit does not make sense in this type of configuration. However, if we consider the values of {$x$} that are greater than -1, then we find that the values of {$f$} are defined! See, so this means that the one-sided limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} can be considered.\\

Obviously, there are two types of one-sided limits. The first type is when the value of {$x$} approaches from the right of {$a$}. This is so because we are considering the values of {$x$} to the right of {$a$}. This only happens when {$f$} is properly defined on that side. In limit notation, this goes like,

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

'+{$$ \lim_{x \to a^{+}} f(x)=L $$}+'

(:tableend:)

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". Likewise, when "L is the limit of {$f(x)$} as {$x$} approaches a from the left, then we may write it as

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

'+{$$ \lim_{x \to a^{-}} f(x)=L $$}+'

(:tableend:)

The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$} while the symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define those one-sided limits formaly.

\begin{definition}[Right-Hand Side Limit]

Let $f$ be a function defined at every point in some open interval (a,b). Then \textbf{the limit of f(x) as x approaches a from the right is L}, notated as

\begin{equation}

\lim_{x \to a^{+}} f(x)=L

\end{equation}

if for every $\epsilon >0$, no matter how small, there exists a $\delta>0$ such that $|f(x)-L|< \epsilon$ whenever $0<x-a<\delta$.

\end{definition}

\begin{definition}[Left-Hand Side Limit]

Let $f$ be a function defined at every point in some open interval (c,a). Then \textbf{the limit of f(x) as x approaches a from the left is L}, notated as

\begin{equation}

\lim_{x \to a^{-}} f(x)=L

\end{equation}

if for every $\epsilon >0$, no matter how small, there exists a $\delta>0$ such that $|f(x)-L|< \epsilon$ whenever $0<x-a<\delta$.

\end{definition}

\indent

As an important remark, the theorems discussed in the previous article still holds true for one-sided limits where "$x-a$" is replaced by "$x-a^{+}$" or "$x-a^{-}$".

\begin{example}

Evaluate the following limits:

\begin{equation}

\lim_{x \to 1^{+}} \frac{x-1}{\sqrt{x^{2}-1}}

\end{equation}

\end{example}

\indent

Notice that if we follow directly theorems discussed before by substituting directly 1 to the function, it gives zero at the denominator. To avoid these, let us rationalize the denominator. Note that rationalizing denominator is not a standard option in solving one-sided limits. The use of such is on a case to case basis. Well anyway, let's proceed.

\newline{}

\indent

First, let us rationalize the function

\begin{eqnarray}

f(x) & = & \frac{x-1}{\sqrt{x^{2}-1}} \\

& = & \frac{x-1}{\sqrt{(x+1)(x-1)}}\cdot \frac{\sqrt{(x-1)(x+1)}}{\sqrt{(x+1)(x-1)}} \\

& = & \frac{(x-1)\sqrt{x^{2}-1}}{\sqrt{(x+1)^{2}(x-1)^{2}}} \\

& = & \frac{(x-1)\sqrt{x^{2}-1}}{(x+1)(x-1)} \\

& = & \frac{\sqrt{x^{2}-1}}{(x+1)}

\end{eqnarray}

\noindent

We have now succesfully rationalized the denominator. Observe that it is not zero at the denominator when $x$ is replaced by $a$. Thus, we now proceed accordingly.

\begin{eqnarray}

\lim_{x \to 1^{+}} f(x) & = & \lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}{(x+1)} \\

& = & \frac{\sqrt{1^{2}-1}}{1+1} \\

& = & \frac{0}{2} \\

& = & 0

\end{eqnarray}

Thus, we have successfully solved for the limit of $f(x)$. In a broader sense, the solutions and techniques employed in solving two-sided limits are still applicable for one-sided limits, as well as the theorems involved.

\end{document}

%rframe width=250px border='0px solid black' padding=2px%Attach:one_sided_limits.jpg | '''Figure 1: Plot of the function {$f(x)=\sqrt{1+x}$}.'''

On the other hand, one sided limits are also considered here. This kind of limits are important because when there is a discontinuity of the function on either side of $a$, then the two-sided limits is not true at all. It may be true on one side but it can't be on the other side. As an illustration, consider the function {$f(x)=\sqrt{1+x}$} which is plotted in Figure 1.\\

Note that the value of the function is imaginary at points {$x<-1$} as seen in the figure. This corresponds to the left side of the the limit supposing that {$a=-1$}. Hence, {$f(x)$} is not defined on the left side of {$a$}, thus, the two sided limit does not make sense in this type of configuration. However, if we consider the values of {$x$} that are greater than -1, then we find that the values of {$f$} are defined! See, so this means that the one-sided limit {$\lim_{x \to -1^{+}}\sqrt{1+x}$} can be considered.\\

Obviously, there are two types of one-sided limits. The first type is when the value of {$x$} approaches from the right of {$a$}. This is so because we are considering the values of {$x$} to the right of {$a$}. This only happens when {$f$} is properly defined on that side. In limit notation, this goes like,

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

'+{$$ \lim_{x \to a^{+}} f(x)=L $$}+'

(:tableend:)

which is read as "L is the limit of {$f(x)$} as {$x$} approaches a from the right". Likewise, when "L is the limit of {$f(x)$} as {$x$} approaches a from the left, then we may write it as

(:table border=3 cellpadding=3 cellspacing=0 align=center:)

(:cellnr:)

'+{$$ \lim_{x \to a^{-}} f(x)=L $$}+'

(:tableend:)

The symbol "{$x \to a^{+}$}" means that {$x$} approaches {$a$} through values greater than {$a$} while the symbol "{$x \to a^{-}$}" means that {$x$} approaches {$a$} through values less than {$a$}. With all these in mind, let us now define those one-sided limits formaly.

\begin{definition}[Right-Hand Side Limit]

Let $f$ be a function defined at every point in some open interval (a,b). Then \textbf{the limit of f(x) as x approaches a from the right is L}, notated as

\begin{equation}

\lim_{x \to a^{+}} f(x)=L

\end{equation}

if for every $\epsilon >0$, no matter how small, there exists a $\delta>0$ such that $|f(x)-L|< \epsilon$ whenever $0<x-a<\delta$.

\end{definition}

\begin{definition}[Left-Hand Side Limit]

Let $f$ be a function defined at every point in some open interval (c,a). Then \textbf{the limit of f(x) as x approaches a from the left is L}, notated as

\begin{equation}

\lim_{x \to a^{-}} f(x)=L

\end{equation}

if for every $\epsilon >0$, no matter how small, there exists a $\delta>0$ such that $|f(x)-L|< \epsilon$ whenever $0<x-a<\delta$.

\end{definition}

\indent

As an important remark, the theorems discussed in the previous article still holds true for one-sided limits where "$x-a$" is replaced by "$x-a^{+}$" or "$x-a^{-}$".

\begin{example}

Evaluate the following limits:

\begin{equation}

\lim_{x \to 1^{+}} \frac{x-1}{\sqrt{x^{2}-1}}

\end{equation}

\end{example}

\indent

Notice that if we follow directly theorems discussed before by substituting directly 1 to the function, it gives zero at the denominator. To avoid these, let us rationalize the denominator. Note that rationalizing denominator is not a standard option in solving one-sided limits. The use of such is on a case to case basis. Well anyway, let's proceed.

\newline{}

\indent

First, let us rationalize the function

\begin{eqnarray}

f(x) & = & \frac{x-1}{\sqrt{x^{2}-1}} \\

& = & \frac{x-1}{\sqrt{(x+1)(x-1)}}\cdot \frac{\sqrt{(x-1)(x+1)}}{\sqrt{(x+1)(x-1)}} \\

& = & \frac{(x-1)\sqrt{x^{2}-1}}{\sqrt{(x+1)^{2}(x-1)^{2}}} \\

& = & \frac{(x-1)\sqrt{x^{2}-1}}{(x+1)(x-1)} \\

& = & \frac{\sqrt{x^{2}-1}}{(x+1)}

\end{eqnarray}

\noindent

We have now succesfully rationalized the denominator. Observe that it is not zero at the denominator when $x$ is replaced by $a$. Thus, we now proceed accordingly.

\begin{eqnarray}

\lim_{x \to 1^{+}} f(x) & = & \lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}{(x+1)} \\

& = & \frac{\sqrt{1^{2}-1}}{1+1} \\

& = & \frac{0}{2} \\

& = & 0

\end{eqnarray}

Thus, we have successfully solved for the limit of $f(x)$. In a broader sense, the solutions and techniques employed in solving two-sided limits are still applicable for one-sided limits, as well as the theorems involved.

\end{document}