Limit of $\frac{2\sin^2 x}{x^3 \sin(1/x)}$ as $x\to0$

Question

Suppose $$f(x) = \frac{2\sin^2 x}{x^3 \sin{\frac{1}{x}}}$$ Now find $$\lim_{x \to 0} f(x)$$

My try : $\lim_{x \to 0} f(x) = 2(\lim_{x \to 0} \frac{\sin x}{x})^2 \times \lim_{x \to 0}\frac{1}{x \sin{\frac{1}{x}}} = 2 \times \lim_{x \to 0}\frac{1}{x \sin{\frac{1}{x}}} $

And I'm stuck here and I have no idea .

Answer 1

Let $g (x)=x\sin ( \frac {1}{x}) $.

$$\forall x\neq 0 \;\;|x\sin (\frac {1}{x})|\leq |x|$$ thus

$$\lim_{x\to0}x\sin (\frac {1}{x})=0.$$

take $x_n=\frac {1}{\frac {\pi}{2}+2n\pi} $ and $y_n=\frac {1}{\frac {3\pi}{2}+2n\pi} $ we have

$g (x_n)\to 0^+$ and $f (x_n)\to+\infty $ $g (y_n)\to 0^-$ and $f (y_n)\to-\infty $

we conclude that $$\lim_{x\to0}f(x) $$ doesn't exist.