1 Limits 1.3 Finding Limits Analytically 1.5 Limits Involving Infinity

1.4 One Sided Limits

In Section 1.1 we explored the three ways in which limits of functions failed to exist:

(a)

The function approached different values from the left and right,
(b)

The function grows without bound, and
(c)

The function oscillates.

In this section we explore in depth the concepts behind #1 by introducing the one-sided limit. We begin with formal definitions that are very similar to the definition of the limit given in Section 1.2, but the notation is slightly different and “ $x\neq c$ ” is replaced with either “ $x<c$ ” or “ $x>c$ .” We will consider #2 in more detail in Section 1.5.

Definition 1.4.1 One Sided Limits

Left-Hand Limit

Let $I$ be an open interval with right endpoint $c$ , and let $f$ be a function defined on $I$ . The limit of $f(x)$ , as $x$ approaches $c$ from the left, is $L$ , or, the left-hand limit of $f$ at $c$ is $L$ , denoted by

\lim_{x\rightarrow c^{-}}f(x)=L,

means that given any $\epsilon>0$ , there exists $\delta>0$ such that for all $x<c$ , if $\left\lvert x-c\right\rvert<\delta$ , then $\left\lvert f(x)-L\right\rvert<\epsilon$ .

Right-Hand Limit

Let $I$ be an open interval with left endpoint $c$ , and let $f$ be a function defined on $I$ . The limit of $f(x)$ , as $x$ approaches $c$ from the right, is $L$ , or, the right-hand limit of $f$ at $c$ is $L$ , denoted by

\lim_{x\rightarrow c^{+}}f(x)=L,

means that given any $\epsilon>0$ , there exists $\delta>0$ such that for all $x>c$ , if $\left\lvert x-c\right\rvert<\delta$ , then $\left\lvert f(x)-L\right\rvert<\epsilon$ .

Practically speaking, when evaluating a left-hand limit, we consider only values of $x$ “to the left of $c$ ,” i.e., where $x<c$ . The admittedly imperfect notation $x\to c^{-}$ is used to imply that we look at values of $x$ to the left of $c$ . The notation has nothing to do with positive or negative values of either $x$ or $c$ . A similar statement holds for evaluating right-hand limits; there we consider only values of $x$ to the right of $c$ , i.e., $x>c$ . We can use the theorems from previous sections to help us evaluate these limits; we just restrict our view to one side of $c$ .

Watch the video:
One-sided limits from graphs from https://youtu.be/nOnd3SiYZqM

We practice evaluating left and right-hand limits through several examples.

Example 1.4.1 Evaluating one sided limits

Let $\displaystyle f(x)=\begin{cases}2x&0\leq x\leq 1\\ 6-2x&1<x<2\end{cases},$ as shown in Figure 1.4.1. Find each of the following: ^†^†margin: Figure 1.4.1: A graph of $f$ in Example 1.4.1. Λ

(a)

$\displaystyle\lim_{x\to 1^{-}}f(x)$
(b)

$\displaystyle\lim_{x\to 1^{+}}f(x)$
(c)

$\displaystyle\lim_{x\to 1}f(x)$
(d)

$\displaystyle f(1)$
(e)

$\displaystyle\lim_{x\to 0^{+}}f(x)$
(f)

$f(0)$
(g)

$\displaystyle\lim_{x\to 2^{-}}f(x)$
(h)

$f(2)$

SolutionFor these problems, the visual aid of the graph is likely more effective in evaluating the limits than using $f$ itself. Therefore we will refer often to the graph.

(a)

As $x$ goes to 1 from the left, we see that $f(x)$ is approaching the value of 2. Therefore $\displaystyle\lim_{x\to 1^{-}}f(x)=2.$
(b)

As $x$ goes to 1 from the right, we see that $f(x)$ is approaching the value of 4. Recall that it does not matter that there is an “open circle” there; we are evaluating a limit, not the value of the function. Therefore $\displaystyle\lim_{x\to 1^{+}}f(x)=4$ .
(c)

The limit of $f$ as $x$ approaches 1 does not exist, as discussed in the first section. The function does not approach one particular value, but two different values from the left and the right.
(d)

Using the definition and by looking at the graph we see that $f(1)=2$ .
(e)

As $x$ goes to 0 from the right, we see that $f(x)$ is also approaching 0. Therefore $\displaystyle\lim_{x\to 0^{+}}f(x)=0$ . Note we cannot consider a left-hand limit at 0 as $f$ is not defined for values of $x<0$ .
(f)

Using the definition and the graph, $f(0)=0$ .
(g)

As $x$ goes to 2 from the left, we see that $f(x)$ is approaching the value of 2. Therefore $\displaystyle\lim_{x\to 2^{-}}f(x)=2.$
(h)

The graph and the definition of the function show that $f(2)$ is not defined.

Note how the left and right-hand limits were different at $x=1$ . This, of course, causes the limit to not exist. The following theorem states what is fairly intuitive: the limit exists precisely when the left and right-hand limits are equal.

Theorem 1.4.1 Limits and One Sided Limits

Let $f$ be a function defined on an open interval $I$ containing $c$ , except possibly at $c$ . Then

\lim_{x\to c}f(x)=L

if, and only if,

\lim_{x\to c^{-}}f(x)=L\quad\text{and}\quad\lim_{x\to c^{+}}f(x)=L.

The phrase “if, and only if” means the two statements are equivalent: they are either both true or both false. If the limit equals $L$ , then the left and right hand limits both equal $L$ . If the limit is not equal to $L$ , then at least one of the left and right-hand limits is not equal to $L$ (it may not even exist). One thing to consider in Examples 1.4.1 – 1.4.4 is that the value of the function may or may not be equal to the value(s) of its left- or right-hand limits, even when these limits agree.

Example 1.4.2 Evaluating limits of a piecewise-defined function

Let $f(x)=\begin{cases}2-x&0<x<1\\ (x-2)^{2}&1<x<2\end{cases},$ as shown in Figure 1.4.2. Evaluate the following.

(a)

$\displaystyle\lim_{x\to 1^{-}}f(x)$
(b)

$\displaystyle\lim_{x\to 1^{+}}f(x)$
(c)

$\displaystyle\lim_{x\to 1}f(x)$
(d)

$\displaystyle f(1)$
(e)

$\displaystyle\lim_{x\to 0^{+}}f(x)$
(f)

$f(0)$
(g)

$\displaystyle\lim_{x\to 2^{-}}f(x)$
(h)

$f(2)$

^†^†margin: Figure 1.4.2: A graph of

f

from Example 1.4.2 Λ

SolutionAgain we will evaluate each using both the definition of $f$ and its graph.

(a)

As $x$ approaches 1 from the left, we see that $f(x)$ approaches 1. Therefore $\displaystyle\lim_{x\to 1^{-}}f(x)=1.$
(b)

As $x$ approaches 1 from the right, we see that again $f(x)$ approaches 1. Therefore $\displaystyle\lim_{x\to 1+}f(x)=1$ .
(c)

The limit of $f$ as $x$ approaches 1 exists and is 1, as $f$ approaches 1 from both the right and left. Therefore $\displaystyle\lim_{x\to 1}f(x)=1$ .
(d)

$f(1)$ is not defined. Note that 1 is not in the domain of $f$ as defined by the problem, which is indicated on the graph by an open circle when $x=1$ .
(e)

As $x$ goes to 0 from the right, $f(x)$ approaches 2. So $\displaystyle\lim_{x\to 0^{+}}f(x)=2$ .
(f)

$f(0)$ is not defined as $0$ is not in the domain of $f$ .
(g)

As $x$ goes to 2 from the left, $f(x)$ approaches 0. So $\displaystyle\lim_{x\to 2^{-}}f(x)=0$ .
(h)

$f(2)$ is not defined as 2 is not in the domain of $f$ .

Example 1.4.3 Evaluating limits of a piecewise-defined function

Suppose $f(x)=\begin{cases}(x-1)^{2}&0\leq x\leq 2,x\neq 1\\ 1&x=1\end{cases}$ , as shown in Figure 1.4.3. Evaluate the following. ^†^†margin: Figure 1.4.3: Graphing $f$ in Example 1.4.3 Λ

(a)

$\displaystyle\lim_{x\to 1^{-}}f(x)$
(b)

$\displaystyle\lim_{x\to 1^{+}}f(x)$
(c)

$\displaystyle\lim_{x\to 1}f(x)$
(d)

$f(1)$

SolutionIt is clear by looking at the graph that both the left and right-hand limits of $f$ , as $x$ approaches 1, is 0. Thus it is also clear that the limit is 0; i.e., $\displaystyle\lim_{x\to 1}f(x)=0$ . It is also clearly stated that $f(1)=1$ .

Example 1.4.4 Evaluating limits of a piecewise-defined function

Let $f(x)=\begin{cases}x^{2}&0\leq x\leq 1\\ 2-x&1<x\leq 2\end{cases}.$ Evaluate the following.

(a)

$\displaystyle\lim_{x\to 1^{-}}f(x)$
(b)

$\displaystyle\lim_{x\to 1^{+}}f(x)$
(c)

$\displaystyle\lim_{x\to 1}f(x)$
(d)

$f(1)$

SolutionIn this example, we will evaluate the limit by only considering the definition of $f$ .

(a)

As $x$ approaches 1 from the left, $f(x)$ is defined to be $x^{2}$ . Therefore

$\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}x^{2}=1.$
(b)

As $x$ approaches 1 from the right, $f(x)$ is defined to be $2-x$ . Therefore

$\lim_{x\to 1+}f(x)=\lim_{x\to 1+}(2-x)=1.$
(c)

Since the right and left hand limits are equal at $x=1$ , i.e., $\displaystyle\lim_{x\to 1^{-}}f(x)=\displaystyle\lim_{x\to 1^{+}}f(x)=1$ , this tells us $\displaystyle\lim_{x\to 1}f(x)=1$ .
(d)

To find $f(1)$ , we use the $x^{2}$ piece of our function, so $f(1)=1$ .

Example 1.4.5 Evaluating limits of an absolute value function

Let $f(x)=\dfrac{\left\lvert x-1\right\rvert}{x-1}.$ Evaluate the following.

(a)

$\displaystyle\lim_{x\to 1^{-}}f(x)$
(b)

$\displaystyle\lim_{x\to 1^{+}}f(x)$
(c)

$\displaystyle\lim_{x\to 1}f(x)$
(d)

$f(1)$

SolutionWe begin by rewriting $\left\lvert x-1\right\rvert$ as a piecewise function.

\left\lvert x-1\right\rvert=\begin{cases}x-1&x\geq 1\\ -(x-1)&x\leq 1\end{cases}

(a)

$\displaystyle\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}\frac{-(x-1)}{x-1}=\lim_{x% \to 1^{-}}-1=-1$
(b)

$\displaystyle\lim_{x\to 1^{+}}f(x)=\lim_{x\to 1^{+}}\frac{x-1}{x-1}=\lim_{x\to 1% ^{+}}1=1$
(c)

$\displaystyle\lim_{x\to 1}f(x)$ does not exist because the left and right hand limits are not equal.
(d)

$f(1)$ is undefined.

In Examples 1.4.1 – 1.4.5 we were asked to find both $\displaystyle\lim_{x\to 1}f(x)$ and $f(1)$ . Consider the following table:

	$\displaystyle\lim_{x\to 1}f(x)$	$f(1)$
Example 1.4.1	does not exist	2
Example 1.4.2	1	not defined
Example 1.4.3	0	1
Example 1.4.4	1	1
Example 1.4.5	does not exist	not defined

Only in Example 1.4.4 do both the function and the limit exist and agree. This seems “nice;” in fact, it seems “normal.” This is in fact an important situation which we explore in Section 1.6, entitled “Continuity.” In short, a continuous function is one in which when a function approaches a value as $x\rightarrow c$ (i.e., when $\displaystyle\lim_{x\to c}f(x)=L$ ), it actually attains that value at $c$ . Such functions behave nicely as they are very predictable.

In the next section we examine one more aspect of limits: limits that involve infinity.

Exercises 1.4

Terms and Concepts

1.

What are the three ways in which a limit may fail to exist?
2.

T/F: If $\displaystyle\lim_{x\to 1^{-}}f(x)=5$ , then $\displaystyle\lim_{x\to 1}f(x)=5$
3.

T/F: If $\displaystyle\lim_{x\to 1^{-}}f(x)=5$ , then $\displaystyle\lim_{x\to 1^{+}}f(x)=5$
4.

T/F: If $\displaystyle\lim_{x\to 1}f(x)=5$ , then $\displaystyle\lim_{x\to 1^{-}}f(x)=5$

Problems

In Exercises 5–10., evaluate each expression using the given graph of $f(x)$ .

5.
(a) $\displaystyle\lim_{x\to 1^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 1^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 1}f(x)$ (d) $f(1)$ (e) $\displaystyle\lim_{x\to 0^{-}}f(x)$ (f) $\displaystyle\lim_{x\to 0^{+}}f(x)$
6.
(a) $\displaystyle\lim_{x\to 1^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 1^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 1}f(x)$ (d) $f(1)$ (e) $\displaystyle\lim_{x\to 2^{-}}f(x)$ (f) $\displaystyle\lim_{x\to 2^{+}}f(x)$
7.
(a) $\displaystyle\lim_{x\to 1^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 1^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 1}f(x)$ (d) $f(1)$
8.
(a) $\displaystyle\lim_{x\to 0^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 0^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 0}f(x)$ (d) $f(0)$
9.
(a) $\displaystyle\lim_{x\to-2^{-}}f(x)$ (b) $\displaystyle\lim_{x\to-2^{+}}f(x)$ (c) $\displaystyle\lim_{x\to-2}f(x)$ (d) $f(-2)$ (e) $\displaystyle\lim_{x\to 2^{-}}f(x)$ (f) $\displaystyle\lim_{x\to 2^{+}}f(x)$ (g) $\displaystyle\lim_{x\to 2}f(x)$ (h) $f(2)$
10.
Let $-3\leq a\leq 3$ be an integer. (a) $\displaystyle\lim_{x\to a^{-}}f(x)$ (b) $\displaystyle\lim_{x\to a^{+}}f(x)$ (c) $\displaystyle\lim_{x\to a}f(x)$ (d) $f(a)$

In Exercises 11–14., evaluate the given limit.

11.

$\displaystyle\lim_{x\to 7^{-}}\sqrt{x^{2}-49}$ .
12.

$\displaystyle\lim_{x\to 5^{+}}\sqrt{x^{2}-25}$ .
13.

$\displaystyle\lim_{x\to-10^{-}}\sqrt{100-x^{2}}$ .
14.

$\displaystyle\lim_{x\to-8^{+}}\sqrt{x^{2}-64}$ .

In Exercises 15–24., evaluate the given limits of the piecewise defined functions $f$ .

15.
$\displaystyle f(x)=\begin{cases}x+1&x\leq 1\\ x^{2}-5&x>1\end{cases}$ (a) $\displaystyle\lim_{x\to 1^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 1^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 1}f(x)$ (d) $f(1)$
16.
$f(x)=\begin{cases}2x^{2}+5x-1&x<0\\ \sin x&x\geq 0\end{cases}$ (a) $\displaystyle\lim_{x\to 0^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 0^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 0}f(x)$ (d) $f(0)$
17.
$\displaystyle f(x)=\begin{cases}x^{2}-1&x<-1\\ x^{3}+1&-1\leq x\leq 1\\ x^{2}+1&x>1\end{cases}$ (a) $\displaystyle\lim_{x\to-1^{-}}f(x)$ (b) $\displaystyle\lim_{x\to-1^{+}}f(x)$ (c) $\displaystyle\lim_{x\to-1}f(x)$ (d) $f(-1)$ (e) $\displaystyle\lim_{x\to 1^{-}}f(x)$ (f) $\displaystyle\lim_{x\to 1^{+}}f(x)$ (g) $\displaystyle\lim_{x\to 1}f(x)$ (h) $f(1)$
18.
$\displaystyle f(x)=\begin{cases}\cos x&x<\pi\\ \sin x&x\geq\pi\end{cases}$ (a) $\displaystyle\lim_{x\to\pi^{-}}f(x)$ (b) $\displaystyle\lim_{x\to\pi^{+}}f(x)$ (c) $\displaystyle\lim_{x\to\pi}f(x)$ (d) $f(\pi)$
19.
$\displaystyle f(x)=\begin{cases}1-\cos^{2}x&x<a\\ \sin^{2}x&x\geq a\end{cases},$ where $a$ is a real number. (a) $\displaystyle\lim_{x\to a^{-}}f(x)$ (b) $\displaystyle\lim_{x\to a^{+}}f(x)$ (c) $\displaystyle\lim_{x\to a}f(x)$ (d) $f(a)$
20.
$\displaystyle f(x)=\begin{cases}x+1&x<1\\ 1&x=1\\ x-1&x>1\end{cases}$ (a) $\displaystyle\lim_{x\to 1^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 1^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 1}f(x)$ (d) $f(1)$
21.
$\displaystyle f(x)=\begin{cases}x^{2}&x<2\\ x+1&x=2\\ -x^{2}+2x+4&x>2\end{cases}$ (a) $\displaystyle\lim_{x\to 2^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 2^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 2}f(x)$ (d) $f(2)$
22.
$\displaystyle f(x)=\begin{cases}a(x-b)^{2}+c&x<b\\ a(x-b)+c&x\geq b\end{cases},$ where $a$ , $b$ and $c$ are real numbers. (a) $\displaystyle\lim_{x\to b^{-}}f(x)$ (b) $\displaystyle\lim_{x\to b^{+}}f(x)$ (c) $\displaystyle\lim_{x\to b}f(x)$ (d) $f(b)$
23.
$\displaystyle f(x)=\begin{cases}\frac{\left\lvert x\right\rvert}{x}&x\neq 0\\ 0&x=0\end{cases}$ (a) $\displaystyle\lim_{x\to 0^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 0^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 0}f(x)$ (d) $f(0)$
24.
$\displaystyle f(x)=\frac{\left\lvert 4-x\right\rvert}{x-4}$ (a) $\displaystyle\lim_{x\to 4^{-}}f(x)$ (b) $\displaystyle\lim_{x\to 4^{+}}f(x)$ (c) $\displaystyle\lim_{x\to 4}f(x)$ (d) $f(4)$

In Exercises 25–28., sketch the graph of a function $f$ that satisfies all of the given conditions.

25.

$\displaystyle\lim_{x\to 1^{-}}f(x)=2$ , $\displaystyle\lim_{x\to 1^{+}}f(x)=-3$ , $f(1)=0$ .
26.

$\displaystyle\lim_{x\to-1}f(x)=3$ , $\displaystyle\lim_{x\to 3^{-}}f(x)=1$ , $\displaystyle\lim_{x\to 3^{+}}f(x)=-2$ , $f(-1)=1$ , $f(3)=-2$ .
27.

$\displaystyle\lim_{x\to-2^{-}}f(x)=1$ , $\displaystyle\lim_{x\to-2^{+}}f(x)=0$ , $\displaystyle\lim_{x\to 0^{-}}f(x)=3$ , $\displaystyle\lim_{x\to 0^{+}}f(x)=-1$ , $f(-2)=4$ , $f(0)=-3$ .
28.

$\displaystyle\lim_{x\to 0^{-}}f(x)=0$ , $\displaystyle\lim_{x\to 0^{+}}f(x)=2$ , $\displaystyle\lim_{x\to 4^{-}}f(x)=-2$ , $\displaystyle\lim_{x\to 4^{+}}f(x)=1$ , $f(0)=2$ , $f(4)=-2$ .

Review

29.

Evaluate the limit: $\displaystyle\lim_{x\to-1}\frac{x^{2}+5x+4}{x^{2}-3x-4}$ .
30.

Evaluate the limit: $\displaystyle\lim_{x\to-4}\frac{x^{2}-16}{x^{2}-4x-32}$ .
31.

Evaluate the limit: $\displaystyle\lim_{h\to 0}\frac{\sqrt{3+h}-\sqrt{3}}{h}$ .
32.

Approximate the limit numerically: $\displaystyle\lim_{h\to 0}\frac{(2+h)^{2}-4}{h}$ .
33.

Approximate the limit numerically: $\displaystyle\lim_{x\to 0.2}\frac{x^{2}+5.8x-1.2}{x^{2}-4.2x+0.8}$ .