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I'm having some trouble intuitively understanding the following limit calculation:

$$ \lim_{x\to 0} x^2\sin\left(\frac{1}{x}\right) = 0 $$

$x^2$ obviously goes to $0$ when $x$ approaches zero but $\sin(\frac{1}{x})$ is undefined for x approaching zero. Is $x^2$ dominant in this case? what about the general case where you have a composition of differentiable and continuous functions where the limit of one of them is undefined. How do you approach that problem?

Thank you for your help in advance

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    $\left|\sin \frac{1}{x} \right| \leq 1$ and hence $x^2 \left|\sin \frac{1}{x} \right| \leq x^2$. It is now clear that the limit is 0. – 2017-02-21
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    @Muralidharan Oh shoot, I hadn't seen your comment. I added this as an answer about 27 seconds later. – 2017-02-21
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    This one can be done by squeeze theorem. The other category (composition of functions) needs to be addressed more carefully. Can you give an example of the same? – 2017-02-21

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Note that $ |\sin x| \le 1$. So we have $$ \bigg{|}x^2 \sin \frac{1}{x}\bigg{|} \le x^2$$ So $$0\le\lim_{x \to 0} \bigg{|}x^2 \sin \frac{1}{x}\bigg{|}\le \lim_{x \to 0} x^2=0 \implies \lim_{x \to 0} x^2 \sin \frac{1}{x}=0$$

$x^2$ effects the limit more because the $\sin$ function is bounded between $-1$ and $1$.

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    Thank you. this pretty much explains it :) – 2017-02-21
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    @user3047143 You're Welcome. – 2017-02-21