I need help to determine the $\lim_{x\to0^-} \frac{x^2\sin(1/x)}{\sin(x)}$?

Question

I need help determining the following limit:

$$\lim_{x\to0^-} \frac{x^2\sin(1/x)}{\sin(x)}$$

Answer 1

Write the function as $$ \frac{x^2 \sin(1/x)}{\sin x} = \frac{x}{\sin x} \cdot x \sin\frac{1}{x} $$ and take the limit of each factor.

Answer 2

$$\lim\limits_{x\to0}\frac{x}{\sin x}=1, \ \lim\limits_{x\to0}x\sin\left(\frac{1}{x}\right)=0$$

Answer 3

Notice that $$\begin{align}\lim\limits_{x\to 0^-}\dfrac{x^2\sin{1/x}}{\sin x}=\lim\limits_{x\to 0^-}\dfrac{\sin{1/x}}{1/x}\cdot\lim\limits_{x\to 0^-}\dfrac{x}{\sin{x}}\\ =\lim\limits_{x\to 0^+}\dfrac{\sin{1/x}}{1/x}\cdot\lim\limits_{x\to 0^+}\dfrac{x}{\sin{x}}\end{align}$$

Let $y=1/x$. Then $x\to 0^+$ iff $ y\to \infty$ and we have $\lim\limits_{x\to 0^+}\dfrac{\sin{1/x}}{1/x}=\lim\limits_{y\to\infty}\dfrac{\sin y}{y}=0.$ Since $$\lim\limits_{x\to 0^+}\dfrac{x}{\sin{x}}=\lim\limits_{x\to 0^+}\dfrac{1}{\frac{\sin x}{x}}=1,$$ the desired limit is $0.$