Question

plot y = \frac{\sin mx}{\sin nx} for m = 2 and n = 5. choose the correct graph below. a. b. c. d. \lim_{x\to0}\frac{\sin 3x}{\sin 2x}=\frac{3}{2} the limit does not exist.

Explanation:

Step1: Recall limit formula

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. For $y = \frac{\sin mx}{\sin nx}$ with $m = 2$ and $n=5$, we can rewrite it as $y=\frac{\sin mx}{mx}\cdot\frac{nx}{\sin nx}\cdot\frac{m}{n}$.

Step2: Calculate the limit

As $x
ightarrow0$, $\lim_{x
ightarrow0}\frac{\sin mx}{mx}=1$ and $\lim_{x
ightarrow0}\frac{nx}{\sin nx}=1$. So $\lim_{x
ightarrow0}\frac{\sin mx}{\sin nx}=\frac{m}{n}=\frac{2}{5}$.
The function $y = \frac{\sin 2x}{\sin 5x}$ has vertical - asymptotes when $\sin 5x = 0$ and $\sin 2x
eq0$. $\sin 5x=0$ when $5x = k\pi,k\in\mathbb{Z}$, i.e., $x=\frac{k\pi}{5},k\in\mathbb{Z}$. And it has a horizontal - asymptote $y = 0$ as $x
ightarrow\pm\infty$. Also, at $x = 0$, the limit is $\frac{2}{5}$. Analyzing the behavior of the sine - function based functions, we can eliminate options based on the number of oscillations and the value at $x = 0$.

Answer:

We need to analyze the graphs one - by - one based on the properties of the function $y=\frac{\sin 2x}{\sin 5x}$ such as its limit at $x = 0$ (which is $\frac{2}{5}$), its vertical asymptotes (when $\sin 5x=0$ and $\sin 2x
eq0$), and its general oscillatory behavior. Without seeing the actual graphs, we know that the function has a well - defined limit of $\frac{2}{5}$ at $x = 0$, and it has multiple vertical asymptotes due to the denominator $\sin 5x = 0$. We look for a graph that has a value of $\frac{2}{5}$ at $x = 0$ and appropriate vertical asymptotes. Since the problem does not provide the content of the graphs for a definite answer, we have completed the analysis of the function's properties. If we assume we are choosing based on the limit at $x = 0$ and general shape, we would look for a graph that approaches $\frac{2}{5}$ as $x
ightarrow0$.