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Limit of x sin(1/x) when x goes to 0? Here when x goes to 0 then 1/x goes to infinity. Then what would be the limit of sin(1/x)? Can you give me the answer?

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    I think this question have been already solve. but anyway the limit of $\sin{\frac{1}{x}}$ does not contribute since the absolute value of this function is bounded by 1. therefore your function $|x\sin{\frac{1}{x}}|\le |x|\to 0 ,~~x\to 0$ – 2017-02-07
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    Just a hint: $\sin 1/x$ has no limit, but it's bounded. – 2017-02-07
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    Then do we have to use a constant k as the limit of sin(1/x) when x goes to 0?- 1<=k<=1 – 2017-02-07

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The limit $$\lim_{x\to 0} x\sin \frac{1}{x}$$ exists since $-1 \leq \sin x \leq 1$ for all $x\in \mathbf R$. The limit $$\lim_{x\to 0} \sin \frac{1}{x}$$ does not exist since we can pick sequences $(x_n)_n$ and $(y_n)_n$ with $x_n =\frac{1}{\pi n}$ and $y_n= \frac{2}{\pi n}$ where $\sin x_n = 0$ for all $n\in \mathbf N$ and $\sin y_n = 1$ for all $n\in \mathbf N$ and therefore $ \lim_{n\to \infty }\sin x_n = 0$ and $\lim_{n\to \infty }\sin y_n = 1$.