Subject: Calculus

One Sided Limits

In the 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.

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


Figure 1: Plot of the function f(x). (Source)

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.

The Two Types of One-Sided Limits

Right-Handed Limit

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,

\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. Knowing this, we can now define a right-handed limit.

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:

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

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] .

Left-Handed Limit

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

\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:

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

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

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] .

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^{-}".



Example #1

Evaluate the following limit:

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

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.
First, let us rationalize the function

f(x) = \frac{x-1}{\sqrt{x^{2}-1}}
 = \frac{x-1}{\sqrt{ (x+1)(x-1)}} * {\sqrt{ (x-1) (x+1)}\over {\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.

\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1^{+}} \frac{\sqrt{x^{2}-1}}\over{ (x+1) }
 =\frac{\sqrt{1^{2}-1}}{1+1}
 =\frac{0}{2}
 =0

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 theorems involved.

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