Question

use a graphing utility to plot y = \frac{\sin mx}{\sin nx} for three different pairs of the given nonzero constants m and n. estimate \lim_{x\to0} \frac{\sin mx}{\sin nx} in each case. then use your work to make a conjecture about the value of \lim_{x\to0} \frac{\sin mx}{\sin nx} for any nonzero values of m and n. plot y = \frac{\sin mx}{\sin nx} for m = 2 and n = 3. choose the correct graph below.

Explanation:

Step1: Recall the limit formula

We know the well - known limit $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. Let's rewrite $\lim_{x
ightarrow0}\frac{\sin mx}{\sin nx}$ as $\lim_{x
ightarrow0}\frac{\sin mx}{mx}\cdot\frac{nx}{\sin nx}\cdot\frac{m}{n}$.

Step2: Apply the limit formula

As $x
ightarrow0$, $mx
ightarrow0$ and $nx
ightarrow0$. So, $\lim_{x
ightarrow0}\frac{\sin mx}{mx} = 1$ and $\lim_{x
ightarrow0}\frac{nx}{\sin nx}=1$. Then $\lim_{x
ightarrow0}\frac{\sin mx}{\sin nx}=\frac{m}{n}$.
When $m = 2$ and $n=3$, $\lim_{x
ightarrow0}\frac{\sin 2x}{\sin 3x}=\frac{2}{3}$.

To graph $y = \frac{\sin mx}{\sin nx}$ for $m = 2$ and $n = 3$, we can use a graphing utility. The function $y=\frac{\sin 2x}{\sin 3x}$ has vertical asymptotes at the values of $x$ for which $\sin 3x=0$ and $\sin 2x
eq0$, i.e., $x = k\frac{\pi}{3},k\in\mathbb{Z},k
eq0$.

The general shape of the graph of $y = \frac{\sin mx}{\sin nx}$ will have a hole or a removable discontinuity at $x = 0$ (since the limit exists), and vertical asymptotes at the roots of $\sin nx$ where $\sin mx
eq0$.

Answer:

The limit $\lim_{x
ightarrow0}\frac{\sin mx}{\sin nx}=\frac{m}{n}$ for non - zero $m$ and $n$. When $m = 2$ and $n = 3$, the limit is $\frac{2}{3}$. The correct graph will have a hole at $x = 0$ and vertical asymptotes at $x=k\frac{\pi}{3},k\in\mathbb{Z},k
eq0$. Without seeing the actual graphs A, B, C, D, we can't choose the specific correct graph among them, but the function has the above - described characteristics.