Question

find the limit using $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$. $lim_{x
ightarrow0}\frac{xcsc2x}{cos9x}$. select the correct choice below and, if necessary, fill in the a. a. $lim_{x
ightarrow0}\frac{xcsc2x}{cos9x}=\frac{1}{2}$ (simplify your answer.) b. the limit does not exist.

Explanation:

Step1: Rewrite csc function

Recall that $\csc(2x)=\frac{1}{\sin(2x)}$. So the limit $\lim_{x
ightarrow0}\frac{x\csc(2x)}{\cos(9x)}=\lim_{x
ightarrow0}\frac{x}{\sin(2x)\cos(9x)}$.

Step2: Manipulate the fraction

Multiply and divide by 2: $\lim_{x
ightarrow0}\frac{x}{\sin(2x)\cos(9x)}=\lim_{x
ightarrow0}\frac{1}{2}\cdot\frac{2x}{\sin(2x)}\cdot\frac{1}{\cos(9x)}$.

Step3: Use limit properties

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$. Let $u = 2x$, as $x
ightarrow0$, $u
ightarrow0$. Also, $\lim_{x
ightarrow0}\cos(9x)=\cos(0) = 1$. So $\lim_{x
ightarrow0}\frac{1}{2}\cdot\frac{2x}{\sin(2x)}\cdot\frac{1}{\cos(9x)}=\frac{1}{2}\cdot\lim_{2x
ightarrow0}\frac{2x}{\sin(2x)}\cdot\lim_{x
ightarrow0}\frac{1}{\cos(9x)}$.

Step4: Evaluate the limits

Since $\lim_{2x
ightarrow0}\frac{2x}{\sin(2x)} = 1$ and $\lim_{x
ightarrow0}\frac{1}{\cos(9x)}=\frac{1}{\cos(0)} = 1$, then $\frac{1}{2}\cdot1\cdot1=\frac{1}{2}$.

Answer:

A. $\lim_{x
ightarrow0}\frac{x\csc(2x)}{\cos(9x)}=\frac{1}{2}$