Question

use $lim_{x
ightarrow0}\frac{sin x}{x}=1$ and/or $lim_{x
ightarrow0}\frac{cos x - 1}{x}=0$ to evaluate the following limit.
$lim_{x
ightarrow0}\frac{sin 6x}{sin 7x}$
select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. $lim_{x
ightarrow0}\frac{sin 6x}{sin 7x}=square$ (type an integer or a simplified fraction.)
b. the limit is undefined.

Explanation:

Step1: Rewrite the limit

We rewrite $\lim_{x
ightarrow0}\frac{\sin6x}{\sin7x}$ as $\lim_{x
ightarrow0}\frac{\sin6x}{6x}\cdot\frac{7x}{\sin7x}\cdot\frac{6}{7}$.

Step2: Apply limit rules

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$. Let $u = 6x$ and $v=7x$. As $x
ightarrow0$, $u
ightarrow0$ and $v
ightarrow0$. So $\lim_{x
ightarrow0}\frac{\sin6x}{6x}=1$ and $\lim_{x
ightarrow0}\frac{7x}{\sin7x}=\lim_{x
ightarrow0}\frac{1}{\frac{\sin7x}{7x}} = 1$.

Step3: Calculate the limit

$\lim_{x
ightarrow0}\frac{\sin6x}{6x}\cdot\frac{7x}{\sin7x}\cdot\frac{6}{7}=1\times1\times\frac{6}{7}=\frac{6}{7}$.

Answer:

A. $\lim_{x
ightarrow0}\frac{\sin6x}{\sin7x}=\frac{6}{7}$