how to find $\lim_{x\to 0}\sin^2(\frac{1}{x})\sin^2 x$

Question

How to find $\lim_{x\to 0}\sin^2(\frac{1}{x})\sin^2 x$ ? I tried using taylor expansion:

$$((x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^5))(\frac{1}{x}-\frac{1}{6x^3}+\frac{1}{120x^5}+O(x^{-5})))^2$$

but it gets very complex. I am looking for simple evalutation. Another option that I tried was using $\sin A\sin B =\frac{1}{2}(\cos (A-B)-\cos (A+B))$, but that too got stuck.

Answer 1

Just notice that

$\qquad\qquad\qquad-1\le\sin t\le1\qquad\implies \qquad 0\le\sin^2\left(\frac{1}{x}\right)\sin^2x\le\sin^2x$

And then use the squeeze theorem.

Answer 2

Squeeze theorem:

$$0\leq \sin^2(\frac{1}{x})\leq 1$$

$$\Rightarrow 0\leq \sin^2(\frac{1}{x})\sin^2(x)\leq \sin^2(x)$$

And $$\lim_{x\rightarrow 0} \sin^2(x) = 0 = \lim_{x\rightarrow 0} 0$$

So

$$\lim_{x\rightarrow 0}\sin^2(\frac{1}{x})\sin^2(x)=0$$