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.
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Hint:
$$0\leq\left|x^k\sin \frac{1}{x^k}\right|\leq\left|x^k\right|\xrightarrow[x\rightarrow0]{}0$$
P..
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1But what about the result that :limit as x approaches zero sinx/x is 1? – Rajath Radhakrishnan Aug 25 '13 at 11:49
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But when $x\rightarrow0, \ \frac1x\rightarrow\infty$ and therefore you cannot use taht result on $\frac{\sin\frac{1}{x}}{\frac1x}$ – P.. Aug 25 '13 at 11:54
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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 $0$.
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$$ \lim_{x\to0} \frac{\sin x}{x} = 1\text{ but } \lim_{x\to0}\frac{\sin (1/x)}{1/x} = 0. $$ The latter equality can be seen by squeezing: $\sin(1/x)$ remains between $1$ and $-1$, so $\dfrac{\sin(1/x)}{1/x}$ remains between $\dfrac{1}{1/x}$ and $\dfrac{-1}{1/x}$, both of which approach $0$.