# Limits of Wildly Oscillating Functions [closed]

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?

## closed as off-topic by RRL, Arnaud D., Ernie060, Leucippus, Daniele TampieriOct 2 at 7:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Ernie060, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

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

• But $xsin\frac{1}{x}$ becomes $0$ in every neighbourhood of $0$ – user679770 Oct 1 at 5:09
• @TobyMak: No, the function is undefined at many other points, $\frac 1{k\pi}$ for integer $k$. – Ross Millikan Oct 1 at 5:19
• 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 $1$. – Toby Mak Oct 1 at 5:20

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

• 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 – user679770 Oct 1 at 5:34
• 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. – user Oct 1 at 5:37
• No I am not using chain rule to find the limits I am talking about the proof of chain rule – user679770 Oct 1 at 5:39
• Do you mean l’Hopital? – user Oct 1 at 5:40
• 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 – user679770 Oct 1 at 5:41