Squeeze Theorem & Trigonometric Limits (College Board AP® Calculus AB): Revision Note

Let $f$ and $g$ be the functions defined by $f\left(x\right)=x^2-6x+13$ and $g\left(x\right)=6x-x^2-5$. It is known that $g\left(x\right)\le f\left(x\right)$ for $0<x<6$.

Let $h$ be a function such that $g\left(x\right)\le h\left(x\right)\le f\left(x\right)$ for $0<x<6$.

Find $\underset{x\to 3}{\lim }h\left(x\right)$, being sure to justify your answer.

Answer:

First find the limits for $f$ and $g$

They are continuous in an open interval containing 3, so you can use substitution

$$\begin{gathered} \underset{x\to 3}{\lim }f\left(x\right)={\left(3\right)}^2-6\left(3\right)+13=4 \\ \underset{x\to 3}{\lim }g\left(x\right)=6\left(3\right)-{\left(3\right)}^2-5=4 \end{gathered}$$

Those are equal, so along with all the other given info this means you can use the squeeze theorem

Be sure to justify your answer by mentioning the squeeze theorem

By the squeeze theorem

$\underset{x\to 3}{\lim }h\left(x\right)=4$

You can use your graphing calculator to check any limit results that you work out analytically.

Find each of the following limits:

(a) $\underset{x\to 0}{\lim }\left(\frac{1-\cos 3x}{x}\right)$

(b) $\underset{x\to 0}{\lim }\left(\frac{\sin 7x}{\sin 4x}\right)$

(c) $\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}\right)$

Answer:

(a)

Substitution would give $\frac{0}{0}$, so instead start with algebraic manipulation

$\frac{1-\cos 3x}{x}=\frac{1-\cos 3x}{x}·\frac{3}{3}=\frac{3\left(1-\cos 3x\right)}{3x}$

Now use properties of limits along with trigonometric limit theorems

We can make use of the result $\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0$

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{3\left(1-\cos 3x\right)}{3x}& =& -3·\underset{x\to 0}{\lim }\frac{\cos 3x-1}{3x}\\ & =& -3·0\end{array}$

$\underset{x\to 0}{\lim }\left(\frac{1-\cos 3x}{x}\right)=0$

(b)

Substitution would give $\frac{0}{0}$, so instead start with algebraic manipulation

$\begin{array}{rcl}\frac{\sin 7x}{\sin 4x}& =& \frac{\sin 7x}{\sin 4x}·\frac{\left({\displaystyle \frac{1}{x}}\right)}{\left({\displaystyle \frac{1}{x}}\right)}·\frac{\left({\displaystyle \frac{7}{7}}\right)}{\left({\displaystyle \frac{4}{4}}\right)}\\ & =& \frac{\left({\displaystyle \frac{7\sin 7x}{7x}}\right)}{\left({\displaystyle \frac{4\sin 4x}{4x}}\right)}\\ & =& \frac{7}{4}·\frac{{\displaystyle \frac{\sin 7x}{7x}}}{{\displaystyle \frac{\sin 4x}{4x}}}\end{array}$

Now use properties of limits along with trigonometric limit theorems

We can make use of the result $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$

$\begin{array}{rcl}\underset{x\to 0}{\lim }\left[\frac{7}{4}·\frac{{\displaystyle \frac{\sin 7x}{7x}}}{{\displaystyle \frac{\sin 4x}{4x}}}\right]& =& \frac{7}{4}·\underset{x\to 0}{\lim }\left[\frac{\left({\displaystyle \frac{\sin 7x}{7x}}\right)}{\left({\displaystyle \frac{\sin 4x}{4x}}\right)}\right]\\ & =& \frac{7}{4}·\frac{\underset{x\to 0}{\lim }\frac{\sin 7x}{7x}}{\underset{x\to 0}{\lim }\frac{\sin 4x}{4x}}\\ & =& \frac{7}{4}·\frac{1}{1}\end{array}$

$\underset{x\to 0}{\lim }\left(\frac{\sin 7x}{\sin 4x}\right)=\frac{7}{4}$

(c)

Substitution would give $\frac{0}{0}$, so instead start with algebraic manipulation

We can make use of the identity ${\sin }^{2}x\equiv 1-{\cos }^{2}x$

$\begin{array}{rcl}\frac{1-\cos x}{x^2}& =& \frac{1-\cos x}{x^2}·\frac{1+\cos x}{1+\cos x}\\ & =& \frac{1-{\cos }^{2}x}{x^2\left(1+\cos x\right)}\\ & =& \frac{{\sin }^{2}x}{x^2\left(1+\cos x\right)}\\ & =& \frac{{\left({\displaystyle \frac{\sin x}{x}}\right)}^2}{1+\cos x}\end{array}$

Now use properties of limits along with trigonometric limit theorems

We can make use of the results $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$

$\begin{array}{rcl}\underset{x\to 0}{\lim }\left[\frac{{\left({\displaystyle \frac{\sin x}{x}}\right)}^2}{1+\cos x}\right]& =& \frac{\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^2}{\underset{x\to 0}{\lim }\left(1+\cos x\right)}\\ & =& \frac{{\left(\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)\right)}^2}{\underset{x\to 0}{\lim }\left(1+\cos x\right)}\\ & =& \frac{1^2}{1+1}\end{array}$

$\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}\right)=\frac{1}{2}$