Question

1. which limit(s) in question #2, are finding the vertical asymptotes of the function? 5. find the limit. you must show how the property $lim_{\theta

ightarrow0}\frac{sin\theta}{\theta}=1$ is used to find the limit. no shortcuts will earn points! a. $lim_{\theta
ightarrow0}\frac{sin(3\theta)}{sin(5\theta)}$ b. $lim_{\theta
ightarrow0}\frac{\tan(5\theta)}{3\theta}$ c. $lim_{\theta
ightarrow0}\frac{sin(3\theta)}{2\theta}$

Explanation:

Step1: Rewrite $\sin(3\theta)$ and $\sin(5\theta)$ for part a

We know that $\sin(A)=A\cdot\frac{\sin(A)}{A}$. So, $\sin(3\theta) = 3\theta\cdot\frac{\sin(3\theta)}{3\theta}$ and $\sin(5\theta)=5\theta\cdot\frac{\sin(5\theta)}{5\theta}$. Then $\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{\sin(5\theta)}=\lim_{\theta
ightarrow0}\frac{3\theta\cdot\frac{\sin(3\theta)}{3\theta}}{5\theta\cdot\frac{\sin(5\theta)}{5\theta}}$.

Step2: Simplify the expression in part a

$\lim_{\theta
ightarrow0}\frac{3\theta\cdot\frac{\sin(3\theta)}{3\theta}}{5\theta\cdot\frac{\sin(5\theta)}{5\theta}}=\lim_{\theta
ightarrow0}\frac{3}{5}\cdot\frac{\frac{\sin(3\theta)}{3\theta}}{\frac{\sin(5\theta)}{5\theta}}$. Using the property $\lim_{x
ightarrow0}\frac{\sin(x)}{x} = 1$, when $\theta
ightarrow0$, we have $\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{3\theta}=1$ and $\lim_{\theta
ightarrow0}\frac{\sin(5\theta)}{5\theta}=1$. So the limit is $\frac{3}{5}\cdot\frac{1}{1}=\frac{3}{5}$.

Step3: Rewrite $\tan(5\theta)$ for part b

Since $\tan(5\theta)=\frac{\sin(5\theta)}{\cos(5\theta)}$, then $\lim_{\theta
ightarrow0}\frac{\tan(5\theta)}{3\theta}=\lim_{\theta
ightarrow0}\frac{\sin(5\theta)}{3\theta\cos(5\theta)}=\lim_{\theta
ightarrow0}\frac{5}{3}\cdot\frac{\sin(5\theta)}{5\theta}\cdot\frac{1}{\cos(5\theta)}$.

Step4: Evaluate the limit in part b

Using $\lim_{\theta
ightarrow0}\frac{\sin(5\theta)}{5\theta}=1$ and $\lim_{\theta
ightarrow0}\cos(5\theta)=\cos(0) = 1$, we get $\lim_{\theta
ightarrow0}\frac{5}{3}\cdot\frac{\sin(5\theta)}{5\theta}\cdot\frac{1}{\cos(5\theta)}=\frac{5}{3}\cdot1\cdot1=\frac{5}{3}$.

Step5: Rewrite $\sin(3\theta)$ for part c

$\sin(3\theta)=3\theta\cdot\frac{\sin(3\theta)}{3\theta}$, so $\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{2\theta}=\lim_{\theta
ightarrow0}\frac{3}{2}\cdot\frac{\sin(3\theta)}{3\theta}$.

Step6: Evaluate the limit in part c

Using $\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{3\theta}=1$, we have $\lim_{\theta
ightarrow0}\frac{3}{2}\cdot\frac{\sin(3\theta)}{3\theta}=\frac{3}{2}\cdot1=\frac{3}{2}$.

Answer:

a. $\frac{3}{5}$
b. $\frac{5}{3}$
c. $\frac{3}{2}$