Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.
$lim_{y
ightarrow0}\frac{sin13y}{6y}$
select the correct choice below and, if necessary, fill in the answer - box in your choice.
a. $lim_{y
ightarrow0}\frac{sin13y}{6y}=square$ (type an integer or a simplified fraction.)
b. the limit does not exist.

Explanation:

Step1: Rewrite the limit

Let $u = 13y$. As $y
ightarrow0$, then $u
ightarrow0$. And $\frac{\sin13y}{6y}=\frac{13}{6}\cdot\frac{\sin13y}{13y}$.

Step2: Apply the limit formula

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. Here $u = 13y$, so $\lim_{y
ightarrow0}\frac{\sin13y}{13y}=1$. Then $\lim_{y
ightarrow0}\frac{\sin13y}{6y}=\frac{13}{6}\lim_{y
ightarrow0}\frac{\sin13y}{13y}$.

Step3: Calculate the limit

Substitute $\lim_{y
ightarrow0}\frac{\sin13y}{13y}=1$ into the above - expression. We get $\frac{13}{6}\times1=\frac{13}{6}$.

Answer:

A. $\lim_{y
ightarrow0}\frac{\sin13y}{6y}=\frac{13}{6}$