Evaluating Limits

Learning Outcomes

Recognize the basic limit laws
Use the limit laws to evaluate the limit of a function

Limit Laws

The first two limit laws were stated earlier in the course and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Basic Limit Results

For any real number $a$ and any constant $c$ ,

$\underset{x\to a}{\lim}x=a$
$\underset{x\to a}{\lim}c=c$

Example: Evaluating a Basic Limit

Evaluate each of the following limits using the basic limit results above.

$\underset{x\to 2}{\lim}x$
$\underset{x\to 2}{\lim}5$

Show Solution

The limit of $x$ as $x$ approaches $a$ is $a$ : $\underset{x\to 2}{\lim}x=2$ .
The limit of a constant is that constant: $\underset{x\to 2}{\lim}5=5$ .

Try It

We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.

Limit Laws

Let $f(x)$ and $g(x)$ be defined for all $x\ne a$ over some open interval containing $a$ . Assume that $L$ and $M$ are real numbers such that $\underset{x\to a}{\lim}f(x)=L$ and $\underset{x\to a}{\lim}g(x)=M$ . Let $c$ be a constant. Then, each of the following statements holds:

Sum law for limits: $\underset{x\to a}{\lim}(f(x)+g(x))=\underset{x\to a}{\lim}f(x)+\underset{x\to a}{\lim}g(x)=L+M$

Difference law for limits: $\underset{x\to a}{\lim}(f(x)-g(x))=\underset{x\to a}{\lim}f(x)-\underset{x\to a}{\lim}g(x)=L-M$

Constant multiple law for limits: $\underset{x\to a}{\lim}cf(x)=c \cdot \underset{x\to a}{\lim}f(x)=cL$

Product law for limits: $\underset{x\to a}{\lim}(f(x) \cdot g(x))=\underset{x\to a}{\lim}f(x) \cdot \underset{x\to a}{\lim}g(x)=L \cdot M$

Quotient law for limits: $\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\dfrac{\underset{x\to a}{\lim}f(x)}{\underset{x\to a}{\lim}g(x)}=\frac{L}{M}$ for $M\ne 0$

Power law for limits: $\underset{x\to a}{\lim}(f(x))^n=(\underset{x\to a}{\lim}f(x))^n=L^n$ for every positive integer $n$ .

Root law for limits: $\underset{x\to a}{\lim}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\lim}f(x)}=\sqrt[n]{L}$ for all $L$ if $n$ is odd and for $L\ge 0$ if $n$ is even

We now practice applying these limit laws to evaluate a limit.

Example: Evaluating a Limit Using Limit Laws

Use the limit laws to evaluate $\underset{x\to -3}{\lim}(4x+2)$ .

Show Solution

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

$\begin{array}{ccccc}\underset{x\to -3}{\lim}(4x+2)\hfill & =\underset{x\to -3}{\lim}4x+\underset{x\to -3}{\lim}2\hfill & & & \text{Apply the sum law.}\hfill \\ & =4 \cdot \underset{x\to -3}{\lim}x+\underset{x\to -3}{\lim}2\hfill & & & \text{Apply the constant multiple law.}\hfill \\ & =4 \cdot (-3)+2=-10\hfill & & & \text{Apply the basic limit results and simplify.}\hfill \end{array}$

Example: Using Limit Laws Repeatedly

Use the limit laws to evaluate $\underset{x\to 2}{\lim}\dfrac{2x^2-3x+1}{x^3+4}$ .

Show Solution

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

$\begin{array}{ccccc}\\ \\ \underset{x\to 2}{\lim}\large \frac{2x^2-3x+1}{x^3+4} & = \large \frac{\underset{x\to 2}{\lim}(2x^2-3x+1)}{\underset{x\to 2}{\lim}(x^3+4)} & & & \text{Apply the quotient law, making sure that} \, 2^3+4\ne 0 \\ & = \large \frac{2 \cdot \underset{x\to 2}{\lim}x^2-3 \cdot \underset{x\to 2}{\lim}x+\underset{x\to 2}{\lim}1}{\underset{x\to 2}{\lim}x^3+\underset{x\to 2}{\lim}4} & & & \text{Apply the sum law and constant multiple law.} \\ & = \large \frac{2 \cdot (\underset{x\to 2}{\lim}x)^2-3 \cdot \underset{x\to 2}{\lim}x+\underset{x\to 2}{\lim}1}{(\underset{x\to 2}{\lim}x)^3+\underset{x\to 2}{\lim}4} & & & \text{Apply the power law.} \\ & = \large \frac{2(4)-3(2)+1}{2^3+4}=\frac{1}{4} & & & \text{Apply the basic limit laws and simplify.} \end{array}$

Try It

Use the limit laws to evaluate $\underset{x\to 6}{\lim}(2x-1)\sqrt{x+4}$ . In each step, indicate the limit law applied.

Hint

Show Solution

$11\sqrt{10}$

Limits of Polynomial and Rational Functions

By now you have probably noticed that, in each of the previous examples, it has been the case that $\underset{x\to a}{\lim}f(x)=f(a)$ . This is not always true, but it does hold for all polynomials for any choice of $a$ and for all rational functions at all values of $a$ for which the rational function is defined.

Limits of Polynomial and Rational Functions

Let $p(x)$ and $q(x)$ be polynomial functions. Let $a$ be a real number. Then,

lim_{x \to a} p (x) = p (a)

$\underset{x\to a}{\lim}p(x)=p(a)$

lim_{x \to a} \frac{p (x)}{q (x)} = \frac{p (a)}{q (a)} when q (a) \neq 0

$\underset{x\to a}{\lim}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)} \, \text{when} \, q(a)\ne 0$

To see that this theorem holds, consider the polynomial $p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0$ . By applying the sum, constant multiple, and power laws, we end up with

\begin{array}{cc} lim_{x \to a} p (x) & = lim_{x \to a} (c_{n} x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{1} x + c_{0}) \\ = c_{n} (lim_{x \to a} x)^{n} + c_{n - 1} (lim_{x \to a} x)^{n - 1} + \dots + c_{1} (lim_{x \to a} x) + lim_{x \to a} c_{0} \\ = c_{n} a^{n} + c_{n - 1} a^{n - 1} + \dots + c_{1} a + c_{0} \\ = p (a) \end{array}

$\begin{array}{cc}\hfill \underset{x\to a}{\lim}p(x)& =\underset{x\to a}{\lim}(c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0)\hfill \\ & =c_n(\underset{x\to a}{\lim}x)^n+c_{n-1}(\underset{x\to a}{\lim}x)^{n-1}+\cdots +c_1(\underset{x\to a}{\lim}x)+\underset{x\to a}{\lim}c_0\hfill \\ & =c_na^n+c_{n-1}a^{n-1}+\cdots +c_1a+c_0\hfill \\ & =p(a)\hfill \end{array}$

It now follows from the quotient law that if $p(x)$ and $q(x)$ are polynomials for which $q(a)\ne 0$ , then

lim_{x \to a} \frac{p (x)}{q (x)} = \frac{p (a)}{q (a)}

$\underset{x\to a}{\lim}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)}$

The example below applies this result.

Example: Evaluating a Limit of a Rational Function

Evaluate the $\underset{x\to 3}{\lim}\dfrac{2x^2-3x+1}{5x+4}$ .

Show Solution

Since 3 is in the domain of the rational function $f(x)=\frac{2x^2-3x+1}{5x+4}$ , we can calculate the limit by substituting 3 for $x$ into the function. Thus,

lim_{x \to 3} \frac{2 x^{2} - 3 x + 1}{5 x + 4} = \frac{10}{19}

$\underset{x\to 3}{\lim}\dfrac{2x^2-3x+1}{5x+4}=\dfrac{10}{19}$

Try It

Evaluate $\underset{x\to -2}{\lim}(3x^3-2x+7)$ .

Hint

Show Solution

$−13$

Watch the following video to see the worked solutions to all examples and try it’s on this page.

Closed Captioning and Transcript Information for Video

Module 2: Limits

Learning Outcomes

Limit Laws

Basic Limit Results

Example: Evaluating a Basic Limit

Try It

Limit Laws

Example: Evaluating a Limit Using Limit Laws

Example: Using Limit Laws Repeatedly

Try It

Limits of Polynomial and Rational Functions

Limits of Polynomial and Rational Functions

Example: Evaluating a Limit of a Rational Function

Try It

Candela Citations