Find the limit of: $$\lim_{x\to 0}\frac{x^2 \sin 1/x}{\sin x}$$
Answer: Since $x^2 \sin\frac{1}{x}$, so $-x^2\leq x^2 \sin\frac{1}{x}\leq x^2$, thus $\lim_{x\to 0}x^2 \sin\frac{1}{x}=0$. I am now stuck with the $\sin x$ under it, how does this affect the limit?
Hint: using your prefered method, prove that $$ \frac {\sin x}x\to 1 $$when $x\to 0$.
details:
once you get that, it remains $$ \left|\frac{x^2\sin 1/x}{\sin x} \right| = \left|\frac{x\sin 1/x}{\sin(x)/x}\right| \le \frac{|x|}{1/2} $$The last inequality becomes true when $x$ is close enough to $0$ because of the definition of th limit, with $\epsilon = 1/2$.
$$\lim_{x\to0}\dfrac{x^2\sin\left(\frac{1}{x}\right)}{\sin x}=\lim_{x\to0}\dfrac{x}{\sin x}{x\sin\left(\frac{1}{x}\right)}$$ Now, assuming existence of $\lim_{x\to0}\dfrac{x}{\sin x}$ and $\lim_{x\to0}{x\sin\left(\frac{1}{x}\right)}$ (you can show they exist), we can use the multiplication law $$\lim_{x\to0}x\sin\left(\frac{1}{x}\right)=\lim_{t\to\infty}\dfrac{\sin\left(t\right)}{t}=0\\ \text{And also }\lim_{x\to0}\dfrac{x}{\sin x}=1$$ Thus, $$\lim_{x\to0}\dfrac{x^2\sin\left(\frac{1}{x}\right)}{\sin x}=0$$
This question is a duplicate. I answered it the other day. See Solve $\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$ .
The easiest approach is to write the expression as $$ \dfrac{\left(\frac{\sin\frac1x}{\frac1x}\right)}{\left(\frac{\sin x}{x}\right)} $$
and recall the behavior of $\frac{\sin x}{x}$ as $x\rightarrow\pm\infty$ (the numerator) and as $x\rightarrow 0$ (the denominator); these are $0$ and $1$, respectively, so the limit is $0/1=\boxed{0}$.
