2
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.