# Prove that the limit $\lim_{x\to 0^+}\frac{1}{x}\sin{(\frac{\pi}{x})}$ does not exist

I have to prove that $$\lim_{x\to 0^+}\frac{1}{x}\sin{\left(\frac{\pi}{x}\right)}$$ does not exist.

My idea: from the definition of function limit, if I found $$x_n\to\infty$$ and $$y_n\to\infty$$ and $$x_n,y_n\neq 0$$ such that $$\frac{1}{x_n}\sin{\left(\frac{\pi}{x_n}\right)}\to l_1$$ and $$\frac{1}{y_n}\sin{\left(\frac{\pi}{y_n}\right)}\to l_2$$ with $$l_1\neq l_2$$ then I have proved the limit does not exist.

I have taken: $$x_n=\frac{1}{2n}$$ and $$y_n=\frac{1}{1/2+2n}$$ and so

$$\lim_{n\to\infty}\frac{1}{x_n}\sin{\left(\frac{\pi}{x_n}\right)}=\lim_{n\to\infty}2n\sin{(2\pi n)}=0=l_1\\ \lim_{n\to\infty}\frac{1}{y_n}\sin{\left(\frac{\pi}{y_n}\right)}=\lim_{n\to\infty}(1/2+2n)\sin{\left(\frac{\pi}{2}+2\pi n\right)}=\infty=l_2$$

Since $$l_1\neq l_2$$ then the limit does not exist.

Question: my work is right?

• Yes, it is fine. Commented Dec 6, 2021 at 9:17
• @KaviRamaMurthy But why for wolfram $l_1$ is indeterminate? wolframalpha.com/input/?i=lim+n-%3Einf+%282+n%29sin%282+pi+n%29 Commented Dec 6, 2021 at 9:20
• @pawel I think WA assumes the variable $n$ is a real value, in which case the limit is indeed indeterminate. But if $n$ is an integer, then the limit is $0$. This is what is meant when Wolfram Alpha says: "Assuming limit refers to a continuous limit | Use the discrete instead". If you click on "discrete", you get: wolframalpha.com/input/…
– 5xum
Commented Dec 6, 2021 at 9:21
• @5xum In my case $n$ is integer since I am using the sequential definition of limit, am I right? Commented Dec 6, 2021 at 9:24
• $l_1=0$ is correct. Commented Dec 6, 2021 at 9:29

Your confusion when using Wolfram Alpha stems from the fact that WA assumes the variable $$n$$ is "continuous", i.e. it can take any real value. In that case,
$$\lim_{n\to\infty} 2n\sin(2\pi n)$$ does not exist.
However, if $$n$$ is an integer, then the limit exists and is $$0$$.