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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Hmm it´s seems like im having problems with lateX coding. – David Jun 19 '17 at 23:26
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I uploaded in a photo because idk where is the error, sorry. – David Jun 19 '17 at 23:28
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1Have you tried the standard 'take the log and apply L'Hospital's rule'? – Simply Beautiful Art Jun 19 '17 at 23:30
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1Did you mean to say the llimit as $x\to\infty$? Otherwise, where does $n$ come in? – Mark Fischler Jun 19 '17 at 23:32
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I think i can´t apply L'Hospital's rule in 1^inf undets. – David Jun 19 '17 at 23:32
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1@David That's why you take the log of it to turn it into $\infty\times0$ or $\frac00$ – Simply Beautiful Art Jun 19 '17 at 23:33
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Yes sorry, i think someone changed, x to infinite – David Jun 19 '17 at 23:33
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@SimplyBeautifulArt But if i use log rule the result i get is infty*1, so i can´t apply L'Hopital's. Maybe i'm wrong, just asking. – David Jun 19 '17 at 23:36
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2How 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 $\infty\times\ln(1)$ – Simply Beautiful Art Jun 19 '17 at 23:38
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@SimplyBeautifulArt God I'm retarded. Sorry i´ve been to many hours studying and my brain is going to explode. Thanks for your patience. – David Jun 19 '17 at 23:40
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2No problem, and you aren't retarded. Helps to write out your work though ;) – Simply Beautiful Art Jun 19 '17 at 23:41
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Now the alternative approach is to use series expansion and the known limit $e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ – Simply Beautiful Art Jun 19 '17 at 23:46
3 Answers
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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 $$ e^{-\pi} $$
Mark Fischler
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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 $\mathrm e^{-\pi}$.
Bernard
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Sorry, how did you do to remove the logarithm in the last part of the exercise? – David Jun 20 '17 at 10:40
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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 $x$. – Bernard Jun 20 '17 at 10:42
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I still do not understand sorry. Are you applying the Taylor series twice? – David Jun 20 '17 at 10:54
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You composed Taylor's expansion for cosine (at order $2$) with the expansion for $\ln(1-x)$ at order $1$, so as to obtain an expansion at order $2$ for the composition $\ln(\cos u)$. Composition of Taylor's series corresponds to composition of functions. – Bernard Jun 20 '17 at 11:00
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Stop thinking L'Hospital's rule is the panacea for determining limits! This most exact and most useful piece of advice, widely shared in the mathematical world, is probably what got this excellent answer two downvotes. If these reactions are not an example of psychological refusal of reality, I do not know what is. – Did Jul 02 '17 at 08:08
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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
$$e^{-\pi}.$$