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I have tried to solve the below limit using the exponential formula and then applying l'Hospital but the problem turns hard to solve. Does anyone knows an easier way for it?

$$\lim_{x\rightarrow\infty}\left(\cos\sqrt{\frac{2\pi}{x}}\right)^x$$

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    $\begingroup$ Have you tried the standard 'take the log and apply L'Hospital's rule'? $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 23:30
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    $\begingroup$ Did you mean to say the llimit as $x\to\infty$? Otherwise, where does $n$ come in? $\endgroup$ – Mark Fischler Jun 19 '17 at 23:32
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    $\begingroup$ @David That's why you take the log of it to turn it into $\infty\times0$ or $\frac00$ $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 23:33
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    $\begingroup$ How do you get $\infty\times1$? Perhaps it may help to write it out? You probably forgot you have to take the log of something in there, so the result should be $\infty\times\ln(1)$ $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 23:38
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    $\begingroup$ No problem, and you aren't retarded. Helps to write out your work though ;) $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 23:41
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Let $$f(x) = x\log\left(\cos\sqrt{\frac{2\pi}{x}}\right) $$ so that your expression is $$ \lim_{x\to\infty} e^{f(x)} = e^{\lim_{x\to\infty} f(x)} $$ Now let $u=1/x$ so that we need to find $$ \lim_{u\to 0} \frac{\log\left(\cos\sqrt{2\pi u}\right)}{u} $$ Now apply l'Hopital and the fact that $\lim_{z\to 0} \tan(az)/z = a$ to get $$ \lim_{u\to 0} \frac{\log\left(\cos\sqrt{2\pi u}\right)}{u} =-\lim_{u\to 0} \frac{\sqrt{\frac\pi2}\tan\sqrt{2\pi u}/\sqrt{u}}{1}=-\pi\sqrt{\frac\pi2}\sqrt{2\pi} $$ giving a result for the original problem of $$ e^{-\pi} $$

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Stop thinking L'Hospital's rule is the panacea for determining limits! Here you simply need Taylor's expansion at order $2$.

First, it is enough to have the limit of the log, $\; x\ln\biggl(\cos\sqrt{\dfrac{2\pi} x}\biggr)$. So set $u= \sqrt{\dfrac{2\pi} x}$: $u$ tends to $0$ when $x$ tends to $\infty$, and \begin{align} x\ln\biggl(\cos\sqrt{\dfrac{2\pi} x}\biggr)&= \dfrac{2\pi}{u^2}\ln(\cos u)=\dfrac{2\pi}{u^2}\ln\Bigl(1-\dfrac{u^2}2+o(u^2)\Bigr)\\&=\dfrac{2\pi}{u^2}\Bigl(-\dfrac{u^2}2+o(u^2)\Bigr)=-\pi +o(1)\to -\pi \end{align} Hence the limit we seek is equal to $\mathrm e^{-\pi}$.

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  • $\begingroup$ Can you explain me where the square root is, please? $\endgroup$ – David Jun 20 '17 at 0:00
  • $\begingroup$ Without forgetting the square root, you would obtain a correct $-\pi$. $\endgroup$ – Yves Daoust Jun 20 '17 at 0:07
  • $\begingroup$ @David: Sorry, I forgot it… I'll update in a moment $\endgroup$ – Bernard Jun 20 '17 at 0:14
  • $\begingroup$ Sorry, how did you do to remove the logarithm in the last part of the exercise? $\endgroup$ – David Jun 20 '17 at 10:40
  • $\begingroup$ I used Taylor's expansion for $\ln(1-x)$: $\;-x-\dfrac{x^2}2-\dfrac{x^3}3-\dotsm$, at order $1$, and substituted $\dfrac{u^2}2$ to $x$. $\endgroup$ – Bernard Jun 20 '17 at 10:42
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$$\cos\sqrt{\frac{2\pi}x}=1-2\sin^2\sqrt{\frac{\pi}{2x}}$$

and

$$x=\frac{\pi\left(\dfrac{\sin\sqrt{\dfrac{\pi}{2x}}}{\sqrt{\dfrac\pi{2x}}}\right)^2}{2\sin^2\sqrt{\dfrac{\pi}{2x}}}.$$

The quotient inside the parenthesis tends to one, and the limit is that of

$$\left(\left(1-2\sin^2\sqrt{\frac{\pi}{2x}}\right)^{1/2\sin^2\sqrt{\frac{\pi}{2x}}}\right)^{\pi(\cdots)^2}$$

or

$$e^{-\pi}.$$

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