Showing $\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=0$ $$\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=\lim_{x\to 0 }\frac{x^2\sin\frac{1}{x}}{x}=\lim_{x\to 0} x\sin\frac{1}{x}=0$$
Is this solution right?
Thank you very much!

A simple proof:
$$ \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x} = \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x} \cdot\frac{x}{x}\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})} = \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x^2\sin(\frac{1}{x})} \cdot x\sin(\frac{1}{x}) $$
The term $\frac{\sin(x^2\sin(\frac{1}{x}))}{x^2\sin(\frac{1}{x})} $ goes to $1$ as $ x \rightarrow 0 $ by the famous identity $\lim_{f(x)\rightarrow0}\frac{\sin(f(x))}{f(x)} = 1 $ as the term $ \sin(1/x) $ is bounded between [-1,1] and the term $ x^2 $ goes to zero, so overall the dummy function $ f(x) = x^2\sin(\frac{1}{x}) \rightarrow 0 $ as $ x \rightarrow 0 $. We now have:
$$ \lim_{x\rightarrow0} x\sin(\frac{1}{x}) = 0$$
as again the term $ \sin(1/x) $ is bounded between [-1,1].
I hope the answer is satisfying.
