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. $lim_{x \to 0}\frac{sin 7x}{17x}$ select the correct choice and, if necessary, fill in the answer box to complete your choice. a. $lim_{x \to 0}\frac{sin 7x}{17x}=$ (type an integer or a simplified fraction.) b. the limit is undefined

Explanation:

Step1: Rewrite the limit

Let $u = 7x$. As $x\to0$, then $u\to0$. We can rewrite $\lim_{x\to0}\frac{\sin7x}{17x}$ as $\lim_{x\to0}\frac{\sin7x}{7x}\cdot\frac{7}{17}$.

Step2: Apply the limit formula

We know that $\lim_{u\to0}\frac{\sin u}{u}=1$. Here $u = 7x$, so $\lim_{x\to0}\frac{\sin7x}{7x}=1$. Then $\lim_{x\to0}\frac{\sin7x}{7x}\cdot\frac{7}{17}=1\cdot\frac{7}{17}$.

Answer:

A. $\lim_{x\to0}\frac{\sin7x}{17x}=\frac{7}{17}$