Subject: Calculus

# Definition Of A Limit

## Calculus.DefinitionOfALimit History

June 26, 2011 by matthew_suan -
Changed line 5 from:
Epsilon and delta are two Greek symbols which are widely used in mathematics and sciences. The convention meaning for epsilon is "small positive infinitesimal quantity" while delta is "difference" or "a change". In this case, the value of {$\delta$} is dependent on {$\epsilon$}. Therefore, if the limit of the function {$f(x)$} as {$x$} approaches {$a$} is {$L$}, then for every positive number {$\epsilon$} , no matter how small it is, there exists a corresponding positive number {$\delta$} such that {$|f(x)-L< \epsilon|$} whenever {$0<|x-a|<\delta$}. This leads to the formal definition of the limit concept:
to:
Epsilon and delta are two Greek symbols which are widely used in [[http://www.mathematics2.com|mathematics]] and sciences. The convention meaning for epsilon is "small positive infinitesimal quantity" while delta is "difference" or "a change". In this case, the value of {$\delta$} is dependent on {$\epsilon$}. Therefore, if the [[Limit of a function|limit of the function]] {$f(x)$} as {$x$} approaches {$a$} is {$L$}, then for every positive number {$\epsilon$} , no matter how small it is, there exists a corresponding positive number {$\delta$} such that {$|f(x)-L< \epsilon|$} whenever {$0<|x-a|<\delta$}. This leads to the formal definition of the limit concept:
June 26, 2011 by matthew_suan -

'''BACK TO''': [[Limits]]
April 23, 2011 by matthew_suan -
Changed line 1 from:
In the [[ Limits | previous article]], we said that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}. This only means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. Since we want {$f$} very close to {$L$} at {$aa}, we make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough. Let's notate this "smallness" by two symbols epsilon ({$\epsilon$}) and delta ({$\delta$}). to: In the [[ Limits | previous article]], we said that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}. This only means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. Since we want {$f$} very close to {$L$} at {$a$}, we make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough. Let's notate this "smallness" by two symbols epsilon ({$\epsilon$}) and delta ({$\delta$}). November 07, 2010 by matthew_suan - Changed line 55 from: We have discussed the definition of a limit and directly apply it in finding limits. With this alone, we can solve and evaluate any limits given. However, solving it using the methods employed above is very tedious. In the previous article on "[[Limit Of A Function | limits of a function]]", we dealt with ways, methods and techniques with the help of several limit theorems to aid us in solving limits. to: We have discussed the definition of a limit and directly applied it in finding limits. With this information alone, we can solve and evaluate any given limits. However, solving using the methods employed above can be very tedious. In the previous article on "[[Limit Of A Function | limits of a function]]", we looked at smarter methods and techniques to find limits with the help of several limit theorems. October 29, 2010 by matthew_suan - Changed line 4 from: %rframe%Attach:deltaepsilon.png to: %rframe border=2%Attach:deltaepsilon.png October 29, 2010 by matthew_suan - Deleted line 0: Changed line 4 from: %rframe%Attach:onesided.jpg to: %rframe%Attach:deltaepsilon.png October 29, 2010 by matthew_suan - Changed lines 1-3 from: !under construction In the [[ Limits | previous article]], we said that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}. This only means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. In other words, we can make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough. Let's notate this "smallness" by two symbols epsilon ({$\epsilon$}) and delta ({$\delta$}). to: In the [[ Limits | previous article]], we said that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}. This only means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. Since we want {$f$} very close to {$L$} at {$aa}, we make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough. Let's notate this "smallness" by two symbols epsilon ({$\epsilon$}) and delta ({$\delta$}).
%rframe%Attach:onesided.jpg
October 28, 2010 by matthew_suan -
Changed lines 1-5 from:
In the [[ Limits | previous article]], we say that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}, means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. In other words, we can make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough.

!!How
are epsilon ({$\epsilon$}) and delta ({$\delta$}) used in limits?
Let's
notate this "smallness" by two symbols: {$\epsilon$} (epsilon) and {$\delta$} (delta). In this case, the value of {$\delta$} is dependent on {$\epsilon$}. Therefore, if the limit of the function {$f(x)$} as {$x$} approaches {$a$} is {$L$}, then for every positive number {$\epsilon$} , no matter how small it is, there exists a corresponding positive number {$\delta$} such that {$|f(x)-L< \epsilon|$} whenever {$0<|x-a|<\delta$}. This leads to the formal definition of the limit concept:
to:
!under construction
In
the [[ Limits | previous article]], we said that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}. This only means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. In other words, we can make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough. Let's notate this "smallness" by two symbols epsilon ({$\epsilon$}) and delta ({$\delta$}).

!!Epsilon
({$\epsilon$}) and delta ({$\delta$}) in limits..
Epsilon
and delta are two Greek symbols which are widely used in mathematics and sciences. The convention meaning for epsilon is "small positive infinitesimal quantity" while delta is "difference" or "a change". In this case, the value of {$\delta$} is dependent on {$\epsilon$}. Therefore, if the limit of the function {$f(x)$} as {$x$} approaches {$a$} is {$L$}, then for every positive number {$\epsilon$} , no matter how small it is, there exists a corresponding positive number {$\delta$} such that {$|f(x)-L< \epsilon|$} whenever {$0<|x-a|<\delta$}. This leads to the formal definition of the limit concept:
Changed line 15 from:
if for every {$\epsilon > 0$} there exists a {$\delta > 0$} such that {$|f(x) - L | < \epsilon$} whenever {$0 < |x-a| < \epsilon$}. Now, using the definition, we will prove that the following function has its limit as indicated.
to:
if for every {$\epsilon > 0$} there exists a {$\delta > 0$} such that {$|f(x) - L | < \epsilon$} whenever {$0 < |x-a| < \delta$}. Now, using the definition, we will prove that the following function has its limit as indicated.
October 28, 2010 by matthew_suan -
Changed line 54 from:
We have discussed the definition of a limit and directly apply it in finding limits. However, this way is very tedious. In the next article on "limits of a function", we will disucss ways, methods and techniques with the help of several limit theorems in solving limits of a function. See you guys!
to:
We have discussed the definition of a limit and directly apply it in finding limits. With this alone, we can solve and evaluate any limits given. However, solving it using the methods employed above is very tedious. In the previous article on "[[Limit Of A Function | limits of a function]]", we dealt with ways, methods and techniques with the help of several limit theorems to aid us in solving limits.
October 28, 2010 by matthew_suan -
October 28, 2010 by matthew_suan -
Changed lines 21-28 from:
$Solution:$ According to the definition, we have to show that for every $\epsilon > 0$ there exists a $\delta > 0$ such that

\begin
{equation}
|(2x-7)-3|< \epsilon \hspace{2mm} \textrm{whenever} \hspace{2mm} \ 0<|x-5|> \delta.
\end{equation}

\noindent
to:
(:toggle id=box2 show='Show Answer with Solutions' init=hide button=1:)
>>id=box2
border='1px solid #999' padding=5px bgcolor=#edf<<
According to the definition, we have to show that for every {$\epsilon > 0$} there exists a {$\delta > 0$} such that
[[<]]
%center%'+{$|(2x-7)-3|< \epsilon$}+' whenever '+{$0<|x-5|> \delta$}+'.
Changed lines 29-30 from:
\begin{equation}
|(2x-7)-3|=|2x-10|=2|x-5|.
to:
%center%'+{$|(2x-7)-3|=|2x-10|=2|x-5|$}+'.

Thus, accordingly, it must be shown that

%center%'+{$2|x-5|< \epsilon$}+' whenever '+{$0<|x-5|> \delta$}+'

which can also be written as

%center%'+{$|x-5|< \frac{\epsilon}{2}$}+' whenever '+{$0<|x-5|> \delta$}+'.

Look at equation (5), seems we have our perfect choice for {$\delta$}. Proceeding accordingly, we choose {$\delta = \frac{1}{2} \epsilon$} and obtain

%center%'+{$2|x-5|< 2 \delta$}+' whenever '+{$0<|x-5|> \delta$}+'
Changed lines 44-71 from:
\noindent
Thus,
according to (3), it must be shown that

\begin
{equation}
2
|x-5|< \epsilon \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta
\end{equation}

\noindent
which can also be written as

\begin{equation}
|x-5|< \frac{\epsilon}{2} \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta.
\end{equation}

\noindent
Look at equation (5), seems we have our perfect choice for $\delta$. Proceeding accordingly, we choose $\delta = \frac{1}{2} \epsilon$ and obtain

\begin{equation}
2|x-5|< 2 \delta \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta
\end{equation}

which is according to (3), this is equal to

\begin{equation}
|(2x-7)-3|< \epsilon \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta.
\end{equation}

\noindent
to:
which is just equal to

%center%'+
{$|(2x-7)-3|< \epsilon$}+' whenever '+{$0<|x-5|> \delta$}+'
Changed lines 50-54 from:
\begin{equation}
\lim_{x \to 5}(2x-7)=3.
\end{equation}

\indent
to:
'+{$$\lim_{x \to 5}(2x-7)=3$$}+'.

>><<
October 28, 2010 by matthew_suan -
In the [[ Limits | previous article]], we say that the limit of {$f$} as {$x$} approaches to a number {$a$} is {$L$}, means that the value of {$f$} can be made closer to {$L$} as we please by taking {$x$} close enough to {$a$}. In other words, we can make the absolute difference of {$f(x)$} and {$L$} ({$|f(x)-L|$}) very small by making the absolute difference of {$x$} and {$a$} ({$|x-a|$}) small enough.

!!How are epsilon ({$\epsilon$}) and delta ({$\delta$}) used in limits?
Let's notate this "smallness" by two symbols: {$\epsilon$} (epsilon) and {$\delta$} (delta). In this case, the value of {$\delta$} is dependent on {$\epsilon$}. Therefore, if the limit of the function {$f(x)$} as {$x$} approaches {$a$} is {$L$}, then for every positive number {$\epsilon$} , no matter how small it is, there exists a corresponding positive number {$\delta$} such that {$|f(x)-L< \epsilon|$} whenever {$0<|x-a|<\delta$}. This leads to the formal definition of the limit concept:

!!Formal Definition of a Limit
Let {$f$} be a function defined at all point {$x$} on the open interval {$I$} that contains {$a$}, except possibly at {$a$} itself. Then, the limit of the function {$f(x)$} as {$x$} approaches {$a$} is {$L$}, which is notated as,

(:table border=3 cellpadding=3 cellspacing=0 align=center:)
(:cellnr:)
'+{$$\lim_{x \to a}f(x)=L$$}+'
(:tableend:)

if for every {$\epsilon > 0$} there exists a {$\delta > 0$} such that {$|f(x) - L | < \epsilon$} whenever {$0 < |x-a| < \epsilon$}. Now, using the definition, we will prove that the following function has its limit as indicated.

!!!Example #1
Using the formal definition of a limit discussed above, verify that

%cframe%'+{$\lim_{x \to 5}(2x-7)=3$}+'.

$Solution:$ According to the definition, we have to show that for every $\epsilon > 0$ there exists a $\delta > 0$ such that

\begin{equation}
|(2x-7)-3|< \epsilon \hspace{2mm} \textrm{whenever} \hspace{2mm} \ 0<|x-5|> \delta.
\end{equation}

\noindent
Now,

\begin{equation}
|(2x-7)-3|=|2x-10|=2|x-5|.
\end{equation}

\noindent
Thus, according to (3), it must be shown that

\begin{equation}
2|x-5|< \epsilon \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta
\end{equation}

\noindent
which can also be written as

\begin{equation}
|x-5|< \frac{\epsilon}{2} \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta.
\end{equation}

\noindent
Look at equation (5), seems we have our perfect choice for $\delta$. Proceeding accordingly, we choose $\delta = \frac{1}{2} \epsilon$ and obtain

\begin{equation}
2|x-5|< 2 \delta \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta
\end{equation}

which is according to (3), this is equal to

\begin{equation}
|(2x-7)-3|< \epsilon \hspace{2mm} \textrm{whenever} \hspace{2mm} 0<|x-5|> \delta.
\end{equation}

\noindent
Thus, as clearly shown and defined,

\begin{equation}
\lim_{x \to 5}(2x-7)=3.
\end{equation}

\indent
We have discussed the definition of a limit and directly apply it in finding limits. However, this way is very tedious. In the next article on "limits of a function", we will disucss ways, methods and techniques with the help of several limit theorems in solving limits of a function. See you guys!