Finding the limit as $n\to\infty$ 
This is how I proceeded with this question but the answer in my book is given to be 0. Is my method correct? Please help so that I can correct myself in case of any mistake.

No notice that 
$$\lim_{x\to0}\frac{\sin x}{x}=1$$
and not $\frac{\sin \frac 1 x}{\frac 1x}$.
Now since $|\sin\frac{1}{x^k}|\leq 1$ then 
$$\lim_{x\to0}x^k\sin\frac{1}{x^k}=0$$
so the given limit is $0$.
