Question

practice exercises
11 - 22. trigonometric limits use theorem 3.10 to evaluate the following limits.

1. $lim_{x

ightarrow0}\frac{sin3x}{x}$

1. $lim_{x

ightarrow0}\frac{sin5x}{3x}$

1. $lim_{x

ightarrow0}\frac{sin7x}{sin3x}$

1. $lim_{x

ightarrow0}\frac{sin3x}{\tan4x}$

1. $lim_{x

ightarrow0}\frac{\tan5x}{x}$

1. $lim_{\theta

ightarrow0}\frac{cos^{2}\theta - 1}{\theta}$

1. $lim_{x

ightarrow0}\frac{\tan7x}{sin x}$

1. $lim_{\theta

ightarrow0}\frac{sec\theta - 1}{\theta}$

1. $lim_{x

ightarrow2}\frac{sin(x - 2)}{x^{2}-4}$

1. $lim_{x

ightarrow - 3}\frac{sin(x + 3)}{x^{2}+8x + 15}$

Explanation:

Step1: Recall the important limit $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
We will use this limit - rule to solve the given problems.

Step2: Solve problem 11

For $\lim_{x
ightarrow0}\frac{\sin3x}{x}$, rewrite it as $\lim_{x
ightarrow0}\frac{\sin3x}{x}=3\lim_{x
ightarrow0}\frac{\sin3x}{3x}$. Let $u = 3x$. As $x
ightarrow0$, $u
ightarrow0$. So $3\lim_{x
ightarrow0}\frac{\sin3x}{3x}=3\times1 = 3$.

Step3: Solve problem 12

For $\lim_{x
ightarrow0}\frac{\sin5x}{3x}$, rewrite it as $\frac{5}{3}\lim_{x
ightarrow0}\frac{\sin5x}{5x}$. Let $u = 5x$. As $x
ightarrow0$, $u
ightarrow0$. So $\frac{5}{3}\lim_{x
ightarrow0}\frac{\sin5x}{5x}=\frac{5}{3}\times1=\frac{5}{3}$.

Step4: Solve problem 13

For $\lim_{x
ightarrow0}\frac{\sin7x}{\sin3x}$, rewrite it as $\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times\frac{3x}{\sin3x}\times\frac{7}{3}$. Let $u_1 = 7x$ and $u_2 = 3x$. As $x
ightarrow0$, $u_1
ightarrow0$ and $u_2
ightarrow0$. So $\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times\frac{3x}{\sin3x}\times\frac{7}{3}=1\times1\times\frac{7}{3}=\frac{7}{3}$.

Step5: Solve problem 14

Recall that $\tan4x=\frac{\sin4x}{\cos4x}$. So $\lim_{x
ightarrow0}\frac{\sin3x}{\tan4x}=\lim_{x
ightarrow0}\frac{\sin3x\cos4x}{\sin4x}=\lim_{x
ightarrow0}\frac{\sin3x}{3x}\times\frac{4x}{\sin4x}\times\frac{3}{4}\times\cos4x$. As $x
ightarrow0$, $\lim_{x
ightarrow0}\frac{\sin3x}{3x}=1$, $\lim_{x
ightarrow0}\frac{4x}{\sin4x}=1$ and $\lim_{x
ightarrow0}\cos4x = 1$. So the limit is $\frac{3}{4}$.

Step6: Solve problem 15

Recall that $\tan5x=\frac{\sin5x}{\cos5x}$. So $\lim_{x
ightarrow0}\frac{\tan5x}{x}=\lim_{x
ightarrow0}\frac{\sin5x}{x\cos5x}=5\lim_{x
ightarrow0}\frac{\sin5x}{5x}\times\frac{1}{\cos5x}$. As $x
ightarrow0$, $\lim_{x
ightarrow0}\frac{\sin5x}{5x}=1$ and $\lim_{x
ightarrow0}\cos5x = 1$. So the limit is $5$.

Step7: Solve problem 16

Recall the double - angle formula $\cos^{2}\theta-1=-\sin^{2}\theta$. So $\lim_{\theta
ightarrow0}\frac{\cos^{2}\theta - 1}{\theta}=\lim_{\theta
ightarrow0}\frac{-\sin^{2}\theta}{\theta}=-\lim_{\theta
ightarrow0}\sin\theta\times\frac{\sin\theta}{\theta}$. As $\theta
ightarrow0$, $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$ and $\lim_{\theta
ightarrow0}\sin\theta = 0$. So the limit is $0$.

Step8: Solve problem 17

Recall that $\tan7x=\frac{\sin7x}{\cos7x}$. So $\lim_{x
ightarrow0}\frac{\tan7x}{\sin x}=\lim_{x
ightarrow0}\frac{\sin7x}{x\cos7x}\times\frac{1}{\sin x}=7\lim_{x
ightarrow0}\frac{\sin7x}{7x}\times\frac{1}{\cos7x}\times\frac{x}{\sin x}$. As $x
ightarrow0$, $\lim_{x
ightarrow0}\frac{\sin7x}{7x}=1$, $\lim_{x
ightarrow0}\cos7x = 1$ and $\lim_{x
ightarrow0}\frac{x}{\sin x}=1$. So the limit is $7$.

Step9: Solve problem 18

Recall that $\sec\theta=\frac{1}{\cos\theta}$. So $\lim_{\theta
ightarrow0}\frac{\sec\theta - 1}{\theta}=\lim_{\theta
ightarrow0}\frac{1 - \cos\theta}{\theta\cos\theta}$. Multiply the numerator and denominator by $1+\cos\theta$: $\lim_{\theta
ightarrow0}\frac{1 - \cos^{2}\theta}{\theta\cos\theta(1 + \cos\theta)}=\lim_{\theta
ightarrow0}\frac{\sin^{2}\theta}{\theta\cos\theta(1 + \cos\theta)}=\lim_{\theta
ightarrow0}\sin\theta\times\frac{\sin\theta}{\theta}\times\frac{1}{\cos\theta(1 + \cos\theta)}$. As $\theta
ightarrow0$, $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$, $\lim_{\theta
ightarrow0}\sin\theta = 0$ and $\lim_{\theta
ightarrow0}\cos\theta = 1$. So the limit is $0$.

Step10: Solve problem 19

Factor $x^{2}-4=(x - 2)(x + 2)$. So $\lim_{x
ightarrow2}\frac{\sin(x - 2)}{x^{2}-4}=\lim_{x
ightarrow2}\frac{\sin(x - 2)}{(x - 2)(x + 2)}$. Let $u=x - 2$. As $x
ightarrow2$, $u
ightarrow0$. So $\lim_{x
ightarrow2}\frac{\sin(x - 2)}{(x - 2)(x + 2)}=\lim_{u
ightarrow0}\frac{\sin u}{u}\times\frac{1}{u + 4}=1\times\frac{1}{4}=\frac{1}{4}$.

Step11: Solve problem 20

Factor $x^{2}+8x + 15=(x + 3)(x+5)$. So $\lim_{x
ightarrow - 3}\frac{\sin(x + 3)}{x^{2}+8x + 15}=\lim_{x
ightarrow - 3}\frac{\sin(x + 3)}{(x + 3)(x + 5)}$. Let $u=x + 3$. As $x
ightarrow - 3$, $u
ightarrow0$. So $\lim_{x
ightarrow - 3}\frac{\sin(x + 3)}{(x + 3)(x + 5)}=\lim_{u
ightarrow0}\frac{\sin u}{u}\times\frac{1}{u+2}=1\times\frac{1}{2}=\frac{1}{2}$.

Answer:

1. $3$
2. $\frac{5}{3}$
3. $\frac{7}{3}$
4. $\frac{3}{4}$
5. $5$
6. $0$
7. $7$
8. $0$
9. $\frac{1}{4}$
10. $\frac{1}{2}$