Question

use $lim_{x \to 0} \frac{sin x}{x}=1$ and/or $lim_{x \to 0} \frac{cos x - 1}{x}=0$ to evaluate the following limit. $lim_{x \to 0} \frac{sin 7x}{\tan x}$ select the correct choice and, if necessary, fill in the answer box to complete your choice. a. $lim_{x \to 0} \frac{sin 7x}{\tan x}=square$ (type an integer or a simplified fraction.) b. the limit is undefined.

Explanation:

Step1: Rewrite tangent

Recall $\tan x=\frac{\sin x}{\cos x}$, so $\lim_{x
ightarrow0}\frac{\sin7x}{\tan x}=\lim_{x
ightarrow0}\frac{\sin7x\cos x}{\sin x}$.

Step2: Use the property $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
Let $u = 7x$. Then $\sin7x=7x\cdot\frac{\sin7x}{7x}$. So $\lim_{x
ightarrow0}\frac{\sin7x\cos x}{\sin x}=\lim_{x
ightarrow0}\frac{7x\cdot\frac{\sin7x}{7x}\cdot\cos x}{\sin x}=7\lim_{x
ightarrow0}\frac{\sin7x}{7x}\cdot\lim_{x
ightarrow0}\frac{x}{\sin x}\cdot\lim_{x
ightarrow0}\cos x$.

Step3: Evaluate each limit

We know $\lim_{x
ightarrow0}\frac{\sin7x}{7x} = 1$, $\lim_{x
ightarrow0}\frac{x}{\sin x}=1$ and $\lim_{x
ightarrow0}\cos x=\cos(0) = 1$.
So $7\lim_{x
ightarrow0}\frac{\sin7x}{7x}\cdot\lim_{x
ightarrow0}\frac{x}{\sin x}\cdot\lim_{x
ightarrow0}\cos x=7\times1\times1\times1 = 7$.

Answer:

A. $\lim_{x
ightarrow0}\frac{\sin7x}{\tan x}=7$