I have a limit as $$\lim_{x\rightarrow 0^{+}} \left((1+\sin\left(\frac{2\pi}{x}\right))\sqrt{x}\right).$$ I am planning to use Squeeze Theorem, so I say that

$-1 \leq \sin(x) \leq 1 \implies -1 \leq \sin(\frac{2\pi}{x}) \leq 1 \implies 0 \leq 1 + \sin\left(\frac{2\pi}{x}\right) \leq 2$

$\implies 0 \leq \left(1 + \sin\left(\frac{2\pi}{x}\right)\right)\sqrt{x} \leq 2\sqrt{x}$

I use the theorem so I get $\lim_{x\rightarrow 0^{+}} ((1+sin(\frac{2\pi}{x})\sqrt{x})=0$.

Is there any problem? Also how can I find the limit $\lim_{x \rightarrow 0^{+}}\sin\left(\cfrac{2\pi}{x}\right)$ so then, I evaluate the whole limit without using Squeeze Theorem.


Is there any problem?

Your derivation is fine. Using the squeeze theorem is a good idea since $$ \lim_{x \rightarrow 0^{+}}\sin\left(\cfrac{2\pi}{x}\right) $$does not exist. Observe that $$ \lim_{n\rightarrow \infty}\sin\left(\cfrac{2\pi}{1/n}\right)=\lim_{n\rightarrow \infty}\sin(2n\pi)=0 $$ and that $$ \lim_{n\rightarrow \infty}\sin\left(\cfrac{2\pi}{1/(n+\frac14)}\right)=\lim_{n\rightarrow \infty}\sin \frac{\pi}2=1\ne0 $$ proving $$ \lim_{x \rightarrow 0^{+}}\sin\left(\cfrac{2\pi}{x}\right) $$ can't exist.

  • $\begingroup$ Unfortunately I could not see how can we conclude that $\lim_{n\rightarrow \infty}\sin\left(\cfrac{2\pi}{1/n}\right)=\lim_{n\rightarrow \infty}\sin(2n\pi)=0$ and $\lim_{n\rightarrow \infty}\sin\left(\cfrac{2\pi}{1/(n+\frac12)}\right)=\lim_{n\rightarrow \infty}\sin(\pi)=-1$. $\endgroup$ – Ninja Mar 4 '17 at 9:31
  • $\begingroup$ Assume $n=1,2,\cdots,$ do you agree with $\sin\left(\cfrac{2\pi}{1/n}\right)=\sin(2n\pi)=0$ and with $\sin\left(\cfrac{2\pi}{1/(n+\frac14)}\right)=\sin(2\pi (n+1/4))=\sin \frac \pi2=1$? $\endgroup$ – Olivier Oloa Mar 4 '17 at 9:46
  • $\begingroup$ If $n$ takes integer values, yes. $\endgroup$ – Ninja Mar 4 '17 at 9:48
  • $\begingroup$ If $x=1/n$ or if $x=1/(n+\frac 14)$, then $x \to 0^+$ as $n \to \infty$. $\endgroup$ – Olivier Oloa Mar 4 '17 at 9:49
  • 1
    $\begingroup$ Yes, that is right. $\endgroup$ – Ninja Mar 4 '17 at 9:58

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