Question

1. (exam 2 review #2b,d) find the following limits

a. 2.5 pts $lim_{\theta
ightarrow0}\frac{cos(\theta)-1}{sin(\theta)}$;
b. 2.5 pts $lim_{x
ightarrow0}csc(x)sin(sin(x))$.

Explanation:

Step1: Recall L - H rule for part A

If we directly substitute $\theta = 0$ into $\frac{\cos(\theta)-1}{\sin(\theta)}$, we get the indeterminate form $\frac{0}{0}$. By L - H rule, if $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}$ is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f'(x)}{g'(x)}$. Let $f(\theta)=\cos(\theta)-1$ and $g(\theta)=\sin(\theta)$. Then $f'(\theta)=-\sin(\theta)$ and $g'(\theta)=\cos(\theta)$.

Step2: Calculate the limit for part A

$\lim_{\theta
ightarrow0}\frac{\cos(\theta)-1}{\sin(\theta)}=\lim_{\theta
ightarrow0}\frac{-\sin(\theta)}{\cos(\theta)}$. Substitute $\theta = 0$ into $\frac{-\sin(\theta)}{\cos(\theta)}$, we have $\frac{-\sin(0)}{\cos(0)}=\frac{0}{1}=0$.

Step3: Rewrite the function for part B

Recall that $\csc(x)=\frac{1}{\sin(x)}$, so $\lim_{x
ightarrow0}\csc(x)\sin(\sin(x))=\lim_{x
ightarrow0}\frac{\sin(\sin(x))}{\sin(x)}$. Let $t = \sin(x)$. As $x
ightarrow0$, $t
ightarrow0$. So the limit becomes $\lim_{t
ightarrow0}\frac{\sin(t)}{t}\cdot\frac{\sin(\sin(x))}{t}\cdot\frac{t}{\sin(x)}$. We know that $\lim_{u
ightarrow0}\frac{\sin(u)}{u}=1$.

Step4: Calculate the limit for part B

$\lim_{x
ightarrow0}\frac{\sin(\sin(x))}{\sin(x)}=\lim_{t
ightarrow0}\frac{\sin(t)}{t}\cdot\lim_{x
ightarrow0}\frac{\sin(\sin(x))}{t}\cdot\lim_{x
ightarrow0}\frac{t}{\sin(x)} = 1$.

Answer:

A. $0$
B. $1$