# Find the Limit (Techniques for Finding Limits)

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

We demonstrate calculating limits using Limit Theorems. We place importance on examples, especially examples with trigonometric and rational functions. We also discuss rationalization, limits of piecewise defined functions, and the Squeeze Theorem.

## Using Limit Theorems

In this topic we concentrate not on the formal definition of a limit of a function of one variable but rather give several examples which emphasis algebra and trigonometry techniques to evaluate limits of functions using basic limit theorems.

Theorem. (Limit Theorems) For any real number $c,$ suppose the functions $f$ and $g$ both have finite limits at $x=c.$ Then

$(1) \quad \displaystyle \lim_{x\to c} k=k$ for any constant $k$

$(2) \quad \displaystyle \lim_{x\to c} x=c$

$(3) \quad \displaystyle \lim_{x\to c}k f(x)=k \lim_{x\to c}f(x)$

$(4) \quad \displaystyle \lim_{x\to c}[f(x)+g(x)]=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)$

$(5) \quad \displaystyle \lim_{x\to c}[f(x)-g(x)]=\lim_{x\to c}f(x)-\lim_{x\to c}g(x)$

$(6) \quad \displaystyle \lim_{x\to c}[f(x)g(x)]=\left( \lim_{x\to c}f(x) \right)\left( \lim_{x\to c}g(x) \right)$

$(7) \quad \displaystyle \lim_{x\to c}[f(x)/g(x)]=\left( \lim_{x\to c}f(x) \right)/\left( \lim_{x\to c}g(x) \right)$

$(8) \quad \displaystyle \lim_{x\to c}[f(x)]^n=\left( \lim_{x\to c}f(x) \right){}^n$

$(9) \quad \displaystyle \lim_{x\to c}P(x)=P(c)$ for any polynomial $P$

$(10) \quad \displaystyle \lim_{x\to c}R(x)=R(c)$ for any rational function $R$

provided all limits exists and $c$ is in the domain of the function.

Example. Find $\displaystyle \lim_{x\to 3}\frac{2x^3-5x+8}{x^2-3}$.

Solution. By using several limit rules, we have \begin{align} \lim_{x\to 3}\frac{2x^3-5x+8}{x^2-3} & =\frac{\lim_{x\to 3}\left(2x^3-5x+8\right)}{\lim_{x\to 3}\left(x^2-3\right)} \\ & =\frac{\lim_{x\to 3}\left(2x^3\right)-\lim_{x\to 3}(5x)+\lim_{x\to 3}(8)}{\lim_{x\to 3}\left(x^2\right)-\lim_{x\to 3}(3)} \\ & =\frac{2 \lim_{x\to 3}\left(x^3\right)- 5\lim_{x\to 3}(x)+8}{\lim_{x\to 3}\left(x^2\right)-3} \\ & =\frac{2 \left(\lim_{x\to 3} x\right){}^3- 5(3)+8}{\left(\lim_{x\to 3}x\right){}^2-3} \\ & =\frac{2 (3)^3- 5(3)+8}{(3)^2-3} =\frac{47}{6} \end{align} Notice this is the same as evaluating the rational function $f(x)$ at $x=3.$

Example. Find $\displaystyle \lim_{x\to 2} \frac{2x^3+x^2-16x+12}{x^2-4}$.

Solution. By using several limit rules, we have \begin{align} \lim_{x\to 2}\frac{2x^3+x^2-16x+12}{x^2-4} & =\lim_{x\to 2}\frac{(x-2)\left(2x^2+5x-6\right)}{(x-2)(x+2)} \\ & =\lim_{x\to 2}\frac{\left(2x^2+5x-6\right)}{(x+2)} \\ & =\frac{\left(2(2)^2+5(2)-6\right)}{((2)+2)} \\ & =3. \end{align} In the previous example notice that we used $$\label{limeq} \lim_{x\to 2}\frac{(x-2)\left(2x^2+5x-6\right)}{(x-2)(x+2)}=\lim_{x\to 2}\frac{\left(2x^2+5x-6\right)}{(x+2)}.$$ Now it is not true that the functions $$f(x)=\frac{\left(2x^2+5x-6\right)}{(x+2)} \qquad \text{and} \qquad g(x)=\frac{\left(2x^2+5x-6\right)}{(x+2)}$$ are the same function because they have different domains. But the above equality is true because $x$ is approaching 2, and not equal to 2. So the point is, because $f(x)=g(x)$ when $x\neq 2$ we can indeed say \eqref{limeq} holds. This is an important part of understanding limits.

Example. Suppose $\displaystyle \lim_{x\to -2^-}f(x)=2,$ $\displaystyle \lim_{x\to -2^+}f(x)=4,$ $\displaystyle \lim_{x\to -2^-}g(x)=0,$ and $\displaystyle \lim_{x\to -2^+}g(x)=0,$ find $$\lim_{x\to -2}[f(x)+g(x)] \quad \text{and} \quad \lim_{x\to -2}[f(x)g(x)].$$

Solution. Since $$\lim_{x\to -2^-}f(x)\neq \lim_{x\to -2^+}f(x)$$ we know $\lim_{x\to -2}f(x)$ does not exist, but this does not imply anything about $\lim_{x\to -2}[f(x)+g(x)]$ nor $\lim_{x\to -2}[f(x)g(x)].$ To find these limits we first find the two one-sided limits, $$\lim_{x\to -2^-}[f(x)+g(x)]=\lim_{x\to -2^-}f(x)+\lim_{x\to -2^-}g(x)=2+0=2$$ $$\lim_{x\to -2^+}[f(x)+g(x)]=\lim_{x\to -2^+}f(x)+\lim_{x\to -2^+}g(x)=4+0=4$$ and since $$\lim_{x\to -2^-}[f(x)+g(x)]\neq \lim_{x\to -2^+}[f(x)+g(x)]$$ we can now say the two-sided limit $\lim_{x\to -2}[f(x)+g(x)]$ does not exist. Similarly, $$\lim_{x\to -2^-}[f(x)g(x)]=\left( \lim_{x\to -2^-}f(x) \right)\left( \lim_{x\to -2^-}g(x)\right)=2(0)=0$$ $$\lim_{x\to -2^+}[f(x)g(x)]=\left( \lim_{x\to -2^+}f(x) \right)\left( \lim_{x\to -2^+}g(x) \right)=4(0)=0$$ and therefore, $\displaystyle \lim_{x\to -2}[f(x)g(x)]=0.$

## Special Trigonometric Limits

Theorem. The following trigonometric limits hold: $$\lim_{x\to 0}\frac{\sin x}{x}=1 \qquad \text{and} \qquad \lim_{x\to 0} \frac{\cos x-1}{x}=0$$

Example. Find $\displaystyle\lim_{x\to 0} \frac{\sin ^2x}{2 x}.$

Solution. We have $$\displaystyle \lim_{x\to 0}\frac{\sin ^2 x}{2x} =\lim_{x\to 0}\frac{\sin x}{2}\frac{\sin x}{x} =0(1)=0.$$ as needed.

Example. Find $\displaystyle \lim_{x\to 0}\frac{\sin x(1-\cos x)}{2 x^2}.$

Solution. We have \begin{align} \lim_{x\to 0}\frac{\sin x(1-\cos x)}{2 x^2} & =\lim_{x\to 0} \frac{1}{2}\frac{\sin x}{x}\frac{1-\cos x}{x} \\ & =\frac{1}{2}\lim_{x\to 0} \left(\frac{\sin x}{x}\right)\lim_{x\to 0}\left(\frac{1-\cos x}{x}\right) \\ & =\frac{1}{2}(1)(0) =0 \end{align} as needed.

Example. Find $\displaystyle\lim_{x\to 0} \frac{\sin 5x}{\sin 4x}.$

Solution. We have \begin{align} \lim_{x\to 0} \frac{\sin 5x}{\sin 4x} & = \lim_{x\to 0} \left(\frac{\sin 5x}{x}\right)\left(\frac{x}{\sin 4x}\right) \\ & =\lim_{x\to 0} \left(\frac{ 5\sin 5x}{5x}\right)\left(\frac{4x}{4 \sin 4x}\right) \\ & =\lim_{x\to 0} \left(\frac{ 5\sin 5x}{5x}\right)\lim_{x\to 0}\left(\frac{4x}{4 \sin 4x}\right) \\ & =\frac{5}{4} \end{align} as needed.

Example. Given the two functions $$g(x)=-27-9 x-6 x^2+x^3+x^4$$ and $$h(x)=21-16 x-3 x^2+2 x^3,$$ find $\lim_{x\to 3} g(x)/h(x)$.

Solution. Try a factor of $(x-3)$ from $g(x)$ obtaining $$g(x)=(x-3)\left(9+6 x+4 x^2+x^3\right)$$ and a graph of $h$ is which also inspires to try to factor of $(x-3)$ from $h(x)$ obtaining $h(x) =(x-3) \left(2x^2+3x-7\right).$ Therefore, \begin{align} \lim_{x\to 3}\frac{-27-9 x-6 x^2+x^3+x^4}{21-16 x-3 x^2+2 x^3} & =\lim_{x\to 3}\frac{(x-3)\left(9+6 x+4 x^2+x^3\right)}{(x-3)\left(2x^2+3x-7\right)} \\ & =\lim_{x\to 3}\frac{9+6 x+4 x^2+x^3}{2x^2+3x-7} \\ & =\frac{9+6 (3)+4 (3)^2+(3)^3}{2(3)^2+3(3)-7} \\ & =\frac{9}{2} \end{align} as needed.

Theorem. If $c$ is a real number, then \begin{array}{cc} \displaystyle \lim_{x\to c} \cos x=\cos c & \displaystyle \lim_{x\to c} \sin x=\sin c \\ \displaystyle \lim_{x\to c} \tan x=\tan c & \displaystyle \lim_{x\to c} \sec x=\sec c \\ \displaystyle \lim_{x\to c} \csc x=\csc c & \displaystyle \lim_{x\to c} \cot x=\cot c. \end{array}

provided $c$ is in the domain of the function.

Example. Find $\displaystyle \lim_{x\to \pi} \frac{x}{\sin x-2 \cos x}$.

Solution. By using several limit rules, we have $$\lim_{x\to \pi }\frac{x}{\sin x-2 \cos x}=\frac{\pi }{\sin \pi -2\cos \pi }=\frac{\pi }{0-2(-1)}=\frac{\pi }{2}$$ as needed.

Example. Find $\displaystyle\lim_{x\to \pi /4}\frac{1-\tan x}{\sin x- \cos x}.$

Solution. We have \begin{align} \lim_{x\to \pi /4}\frac{1-\tan x}{\sin x- \cos x} & =\lim_{x\to \pi /4}\frac{1-\frac{\sin x}{\cos x}}{\sin x- \cos x} \\
& =\lim_{x\to \pi /4}\frac{\frac{\cos x-\sin x}{\cos x}}{\sin x- \cos x} \\ & =\lim_{x\to \pi /4}\frac{\cos x-\sin x}{\cos x}\frac{1}{\sin x- \cos x} \\
&=\lim_{x\to \pi /4}\frac{-1}{\cos x} \\ & =\lim_{x\to \pi /4}\frac{-1}{\frac{1}{\sqrt{2}}} =-\sqrt{2} \end{align} as needed.

Example. Find $\displaystyle \lim_{x\to0}\frac{\tan x}{\sin x}$.

Solution. We have \begin{align} \lim_{x\to 0}\frac{\tan x}{\sin x} & =\lim_{x\to 0} \left(\frac{\sin x}{\cos x}\right)\left(\frac{1}{\sin x}\right) =\lim_{x\to 0} \frac{1}{\cos x} =1 \end{align} as needed.

## Using Rationalization To Find the Limit

Example. Compute $\displaystyle \lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}.$

Solution. We have \begin{align} \lim_{x\to 1}\frac{\sqrt{x}-1}{x-1} & =\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}\frac{\sqrt{x}+1}{\sqrt{x}+1} \\ & =\lim_{x\to 1}\frac{1}{\sqrt{x}+1} \\ & =\frac{1}{\sqrt{1}+1}=\frac{1}{2} \end{align} as needed.

Example. Find $\displaystyle \lim_{x\to\pi}\frac{x-\pi }{\sqrt{x}-\sqrt{\pi }}$.

Solution. We have \begin{align} \lim_{x\to \pi }\frac{x-\pi }{\sqrt{x}-\sqrt{\pi } } & =\lim_{x\to \pi }\left(\frac{x-\pi }{\sqrt{x}-\sqrt{\pi } }\right)\left(\frac{\sqrt{x}+\sqrt{\pi }}{\sqrt{x}+\sqrt{\pi }}\right) \\ & =\lim_{x\to \pi }\frac{(x-\pi )\left(\sqrt{x}+\sqrt{\pi }\right)}{x-\pi } \\ & =2\sqrt{\pi } \end{align} as needed.

Example. Find $\displaystyle\lim_{x\to 1} \frac{x-1}{\sqrt[3]{x}-1}$.

Solution. We have \begin{align} \lim_{x\to 1}\frac{x-1}{\sqrt[3]{x}-1} & =\lim_{x\to 1}\left(\frac{x-1}{\sqrt[3]{x}-1}\right)\left(\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\right) \\
& =\lim_{x\to 1}\frac{(x-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{x-1} \\ & =\lim_{x\to 1}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right) \\ & =3 \end{align} as needed.

## Limits of Piecewise Functions

Example. Find $$\lim_{x\to 0} \begin{cases} 2(x+1) & x<3 \\ 4 & x=3 \\ x^2-1 & x>3. \end{cases}$$

Solution. Since $x\to 0$ we know that $x<3$ and so we use $2(x+1)$ to evaluate the limit of the piecewise function, $$\lim_{x\to 0} \begin{cases} 2(x+1) & x<3 \\ 4 & x=3 \\ x^2-1 & x>3. \end{cases} =\lim_{x\to 0} 2(x+1)=2$$ as needed.

Example. Find $\displaystyle \lim_{x\to 7} f(x)$ given the function f(x)=
\begin{cases} 2\left(x-x^2\right) & x<7 \\ -83 & x=7 \\ (x-7)^2-84 & x>7. \end{cases}

Solution. Since the function is pieced together at $x=7$ we will evaluate two one sided limit. First the limit from the left, we have $$\lim_{x\to 7^-}f(x)=-84$$ and for the limit from the right we have $$\lim_{x\to 7^+}f(x)=-84.$$ Since $$\lim_{x\to 7^-} f(x)=\lim_{x\to 7^+}f(x)$$ we know the two-sided limit must exist and we have $$\lim_{x\to 7}f(x)=-84$$ even though $f(7)=-83.$

## Squeeze Theorem

Next we state the squeeze theorem and through an example show how to use it. Basically, the idea is to bound a function on both sides by functions whose limits can be more easily computed; and thus in the process squeeze the value of the limit of the original function out.

Theorem. (Squeeze Theorem) Let $f, g$ and $h$ be functions of $x$. If both of the following conditions hold

(1) $g(x)\leq f(x)\leq h(x)$ on an open interval containing $c$ and

(2) $\lim_{x\to c}g(x) =\lim_{x\to c}h(x) =L$

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

Example. Use the squeeze rule to find the limit of $\displaystyle f(x)=x \cos \left(\frac{1}{x}\right)$ as $x\to 0^+.$

Solution. We are interested in this function around $x>0$. Knowing that the cosine function is always less than or equal to one, we see that when $-1\leq x\leq 1$ we have $$-x\leq f(x)=x \cos \left(\frac{1}{x}\right)\leq x.$$ Since $\displaystyle \lim_{x\to 0} -x=\lim_{x\to 0} x=0$ we have $\displaystyle \lim_{x\to 0}x \cos \left(\frac{1}{x}\right)=0$ by the squeeze theorem.

## Exercises on Finding the Limit

Exercise. Find the following limits.

$(1) \quad \displaystyle \lim_{x\to -7}(2x+5)$

$(2) \quad \displaystyle \lim_{x\to -2}\left(x^3-2x^2+4x+8\right)$

$(3) \quad \displaystyle \lim_{x\to 2}\left(\frac{x+3}{x+6}\right)$

$(4) \quad \displaystyle \lim_{y\to 2}\left(\frac{y+2}{y^2+5y+6}\right)$

$(5) \quad \displaystyle \lim_{y\to -3}(5-y)^{4/3}$

$(6) \quad \displaystyle \lim_{h\to 0}\left(\frac{5}{\sqrt{5h+4}+2}\right)$

$(7) \quad \displaystyle \lim_{x\to 5}\left(\frac{x-5}{x^2-25}\right)$

$(8) \quad \displaystyle \lim_{x\to 2}\left(\frac{x^2-7x+10}{x-2}\right)$

$(9) \quad \displaystyle \lim_{x\to -2}\left(\frac{-2x-4}{x^3+2x^2}\right)$

$(10) \quad \displaystyle \lim_{x\to 1}\left(\frac{x-1}{\sqrt{x+3}-2}\right)$

$(11) \quad \displaystyle \lim_{x\to -2}\left(\frac{x+2}{\sqrt{x^2+5}-3}\right)$

Exercise. Suppose $\displaystyle \lim_{x\to 4}f(x)=0$ and $\displaystyle \lim_{x\to 4}g(x)=3.$ Using limit laws find the limits

$(1) \quad \displaystyle \lim_{x\to 4}(g(x)+3)$

$(2) \quad \displaystyle \lim_{x\to 4}xf(x)$

$(3) \quad \displaystyle \lim_{x\to 4}(g(x))^2$

$(4) \quad \displaystyle \lim_{x\to 4}\frac{g(x)}{f(x)+1}.$

Exercise. Suppose that $\displaystyle \lim_{x\to -2}p(x)=4,$ $\displaystyle \lim_{x\to -2}r(x)=0,$ and ${\displaystyle \lim_{x\to -2}s(x)=-3.}$ Using limit laws find the limits

$(1) \quad \displaystyle \lim_{x\to -2}(p(x)+r(x)+s(x)),$

$(2) \quad \displaystyle \lim_{x\to -2}p(x)r(x)s(x),$

$(3) \quad \displaystyle \lim_{x\to -2}\left(\frac{-4p(x)+5 r(x)}{s(x)}\right).$

Exercise. Using limit laws evaluate the limit, $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ where $f(x)=x^2$ and $x=1.$

Exercise. Using limit laws evaluate the limit, $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ where $f(x)=3x-4$ and $x=2.$

Exercise. If $\displaystyle \lim_{x\to -2}\frac{f(x)}{x^2}=1,$ find $\lim_{x\to -2}f(x)$ and $\lim_{x\to -2}\frac{f(x)}{x}.$

Exercise. If $\displaystyle \lim_{x\to 2}\frac{f(x)-5}{x-2}=3,$ find $\displaystyle \lim_{x\to 2}f(x).$ Also if $\displaystyle \lim_{x\to 2}\frac{f(x)-5}{x-2}=4$ find $\displaystyle\lim_{x\to 2}f(x).$