# 2.3 The limit laws

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• Recognize the basic limit laws.
• Use the limit laws to evaluate the limit of a function.
• Evaluate the limit of a function by factoring.
• Use the limit laws to evaluate the limit of a polynomial or rational function.
• Evaluate the limit of a function by factoring or by using conjugates.
• Evaluate the limit of a function by using the squeeze theorem.

In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.

## Evaluating limits with the limit laws

The first two limit laws were stated in [link] 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 ,

1. $\underset{x\to a}{\text{lim}}x=a$
2. $\underset{x\to a}{\text{lim}}c=c$

## Evaluating a basic limit

Evaluate each of the following limits using [link] .

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

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

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

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

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

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

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

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

Root law for limits : $\underset{x\to a}{\text{lim}}\sqrt[n]{f\left(x\right)}=\sqrt[n]{\underset{x\to a}{\text{lim}}f\left(x\right)}=\sqrt[n]{L}$ 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.

## Evaluating a limit using limit laws

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

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.

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

## Using limit laws repeatedly

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

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.

$\begin{array}{}\\ \\ \underset{x\to 2}{\text{lim}}\frac{2{x}^{2}-3x+1}{{x}^{3}+4}\hfill & =\frac{\underset{x\to 2}{\text{lim}}\left(2{x}^{2}-3x+1\right)}{\underset{x\to 2}{\text{lim}}\left({x}^{3}+4\right)}\hfill & & & \text{Apply the quotient law, making sure that.}\phantom{\rule{0.2em}{0ex}}{\left(2\right)}^{3}+4\ne 0\hfill \\ & =\frac{2·\underset{x\to 2}{\text{lim}}{x}^{2}-3·\underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{\underset{x\to 2}{\text{lim}}{x}^{3}+\underset{x\to 2}{\text{lim}}4}\hfill & & & \text{Apply the sum law and constant multiple law.}\hfill \\ & =\frac{2·{\left(\underset{x\to 2}{\text{lim}}x\right)}^{2}-3·\underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{{\left(\underset{x\to 2}{\text{lim}}x\right)}^{3}+\underset{x\to 2}{\text{lim}}4}\hfill & & & \text{Apply the power law.}\hfill \\ & =\frac{2\left(4\right)-3\left(2\right)+1}{{\left(2\right)}^{3}+4}=\frac{1}{4}.\hfill & & & \text{Apply the basic limit laws and simplify.}\hfill \end{array}$

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