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Does the limit $$\lim_{x \to 0}\frac{x\sin\frac{1}{x}}{x\sin\frac{1}{x}}$$ exists. If yes then why and if no then why?

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You have to look carefully at your definition of a limit. In the definition I learned, which is the one quoted in Wikipedia, to say $\lim_{x \to 0}f(x)=L$ you must have $\forall \epsilon \exists \delta \ 0 \lt |x| \lt \delta \implies |f(x)-L| \lt \epsilon$ If $f(x)$ fails to be defined at any point in $(-\epsilon, \epsilon)$ this fails. Your function is not defined at any point $x=\frac 1{k \pi}$ for $k$ any integer, so there is no limit at $0$.

If your definition of a limit just refers to points where $f(x)$ is defined, your function will have limit $1$. The fact that the function is not defined at $0$ is immaterial either way. In the question linked by gimusi careful definitions are cited which allow $f(x)$ not to be defined in the interval. There are good arguments for that type of definition, perhaps requiring that $f(x)$ be defined on a set of points converging on the point where you want your limit.

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Hint: $\frac{x \sin\frac{1}{x}}{x \sin\frac{1}{x}}=1$ for all $x$ such that $x \sin \frac{1}{x} \ne 0.$

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  • $\begingroup$ But $xsin\frac{1}{x}$ becomes $0$ in every neighbourhood of $0$ $\endgroup$ – user679770 Oct 1 at 5:09
  • $\begingroup$ @TobyMak: No, the function is undefined at many other points, $\frac 1{k\pi}$ for integer $k$. $\endgroup$ – Ross Millikan Oct 1 at 5:19
  • $\begingroup$ Edited: Nowhere does it approach $0$. Although the function is undefined at $x=0$, and whether we approach from the left or the right side, the value of the limit is still $1$. $\endgroup$ – Toby Mak Oct 1 at 5:20
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Yes it exists! Indeed according to the more general definition of limit, we exclude the points such that $\sin(1/x)=0$ therefore since the ratio is equal to $1$ the limit is $1$ by definition.

Refer also to the strictly related

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  • $\begingroup$ But when we prove chain rule we find the limit of the ratio as $∆x$ tends to $0$ $$(\frac{\Delta y}{\Delta u})(\frac{\Delta u}{\Delta x})$$ and say that going with this argument will lead to wrong result if $u$ is Wildly Oscillating near $0$ There why we don't use this general definition of limit $\endgroup$ – user679770 Oct 1 at 5:34
  • $\begingroup$ Why should we use chain rule to calculate the limit? Refer to the given link carefully. Of course if assume a different definition of limit you can conclude that the limit doesn’t exist but it is, in a certain sense, not satisfactory. For that reason at an height level of study the more general definition apply. $\endgroup$ – user Oct 1 at 5:37
  • $\begingroup$ No I am not using chain rule to find the limits I am talking about the proof of chain rule $\endgroup$ – user679770 Oct 1 at 5:39
  • $\begingroup$ Do you mean l’Hopital? $\endgroup$ – user Oct 1 at 5:40
  • $\begingroup$ No whatbi mean is that in the above question you said that the limit is $1$ from the more general definition of limits then while proving chain rule why don't we use this general definition of limit $\endgroup$ – user679770 Oct 1 at 5:41

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