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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?

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    $\begingroup$ Yes, it is fine. $\endgroup$ Commented Dec 6, 2021 at 9:17
  • $\begingroup$ @KaviRamaMurthy But why for wolfram $l_1$ is indeterminate? wolframalpha.com/input/?i=lim+n-%3Einf+%282+n%29sin%282+pi+n%29 $\endgroup$
    – pawel
    Commented Dec 6, 2021 at 9:20
  • $\begingroup$ @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/… $\endgroup$
    – 5xum
    Commented Dec 6, 2021 at 9:21
  • $\begingroup$ @5xum In my case $n$ is integer since I am using the sequential definition of limit, am I right? $\endgroup$
    – pawel
    Commented Dec 6, 2021 at 9:24
  • $\begingroup$ $l_1=0$ is correct. $\endgroup$ Commented Dec 6, 2021 at 9:29

1 Answer 1

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Yes, your attempt is correct.

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 $0$.


Note that Wolfram Alpha directly explains this:

Explanation

After you click "discrete" in the image above, you get the result you were looking for:

Correct result

which you can also get on this link.

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