MA005: Calculus I

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

THEOREM 2.4

Basic Limit Results

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

i. $\lim\limits_{x \rightarrow a} x=a$                                                                                                                                                                  

ii. $\lim\limits_{x \rightarrow a} c=c$                                                                                                                                                     

EXAMPLE 2.13

Evaluating a Basic Limit

Evaluate each of the following limits using Basic Limit Results.

a. $\lim\limits_{x \rightarrow 2} x$

b. $\lim\limits_{x \rightarrow 2} 5$

Solution

a. The limit of $x$ as $x$ approaches $a$ is a: $\lim\limits_{x \rightarrow 2} x=2$.

b. The limit of a constant is that constant: $\lim\limits_{x \rightarrow 2} 5=5$.

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

THEOREM 2.5

Limit Laws

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

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

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

Constant multiple law for limits: $\lim\limits_{x \rightarrow a} c f(x)=c \cdot \lim\limits_{x \rightarrow a} f(x)=c L$

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

Quotient law for limits: $\lim\limits_{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits_{x \rightarrow a} f(x)}{\lim\limits_{x \rightarrow a} g(x)}=\frac{L}{M}$ for $M \neq 0$

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

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

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

EXAMPLE 2.14

Evaluating a Limit Using Limit Laws

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

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.

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

EXAMPLE 2.15

Using Limit Laws Repeatedly

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

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.

$\begin{aligned} \lim\limits_{x \rightarrow 2} \frac{2 x^{2}-3 x+1}{x^{3}+4} &=\frac{\lim\limits_{x \rightarrow 2}\left(2 x^{2}-3 x+1\right)}{\lim\limits_{x \rightarrow 2}\left(x^{3}+4\right)} \qquad \qquad \text { Apply the quotient law, making sure that. }(2)^{3}+4 \neq 0\ \\ &=\frac{2 \cdot \lim\limits_{x \rightarrow 2} x^{2}-3 \cdot \lim\limits_{x \rightarrow 2} x+\lim\limits_{x \rightarrow 2}}{\lim\limits_{x \rightarrow 2} x^{3}+\lim\limits_{x \rightarrow 2} 4} \qquad \text{Apply the sum law and constant multiple law.} \\ &=\frac{2 \cdot\left(\lim\limits_{x \rightarrow 2} x\right)^{2}-3 \cdot \lim\limits_{x \rightarrow 2} x+\lim\limits_{x \rightarrow 2}}{\left(\lim\limits_{x \rightarrow 2} x\right)^{3}+\lim\limits_{x \rightarrow 2} 4} \qquad \text{Apply the power law.} \\ &=\frac{2(4)-3(2)+1}{(2)^{3}+4}=\frac{1}{4} \qquad \qquad \text{Apply the basic limit laws and simplify.} \end{aligned}$

CHECKPOINT 2.11

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

Source: OpenStax, https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws
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