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Find $$\lim_{n\to \infty}\left(\cos{x\over n}\right)^{n^2}$$ where $x\in \Bbb{R}.$ I tried using taylor series. A complete mess, and an area I am not very good at. I tried using $e$ which also gave me nothing. I am lost a little. I would appreciate your help, thorough or hinted.

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    $\begingroup$ What do you mean by "I tried using $e$"? The number $e$ is not a mathematical technique which can be used to solve problems. $\endgroup$
    – JimmyK4542
    Jan 27 '15 at 21:48
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    $\begingroup$ @JimmyK4542 He is probably referring to $x=e^{\log x}$? $\endgroup$ Jan 27 '15 at 21:52
  • $\begingroup$ You can simplify the limit by taking the natural log of the stuff inside the limit. Then, I believe you can evaluate the resulting limit using one or more applications of L'Hopital's Rule. Finally, take e^(your result) to get your final answer. $\endgroup$
    – nukeguy
    Jan 27 '15 at 22:00
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Taylor expansion should work. Since the argument of the cosine approaches zero in the given limit, we can neglect all higher-order terms and are left with

$\cos \frac{x}{n} \approx 1 - \frac{x^2}{2n^2}$

We can now write

$(cos \frac{x}{n})^{n^2} = \left[ \left(1 - \frac{x^2}{2n^2}\right)^\frac{2n^2}{x^2}\right]^\frac{x^2}{2} \to e^{-\frac{x^2}{2}}$ in the limit of $n\to\infty$

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  • $\begingroup$ I can't make it :( I don't know what $x_0$ I should pick $\endgroup$ Jan 27 '15 at 22:05
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    $\begingroup$ For $n\to\infty$, $x/n\to0$. So $x_0=0$, if I understood your question correctly. $\endgroup$
    – jkard178
    Jan 27 '15 at 22:11
  • $\begingroup$ When I differentiate I got a really big expression. If I pick $n_0=0$ It is not defined... $\endgroup$ Jan 27 '15 at 22:20
  • $\begingroup$ I accidentally used the whole expression oops :$ $\endgroup$ Jan 27 '15 at 22:22
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Combine your two ideas!

This demonstration is not altogether rigorous but it does get the right answer:

Assume $x$ is not a multiple of $\pi$. Taylor expand in $\frac{x}{n}$: $$ \cos \frac{x}{n} = \left( 1 - \frac{x^2}{2n^2} + O(n^{-4}) \right) $$ Then $$ \lim_{n\rightarrow\infty} \left( \cos \frac{x}{n} \right)^{n^2} = \lim_{n\rightarrow\infty} \left(1 - \frac{x^2}{2n^2} \right)^{n^2} = \lim_{n\rightarrow\infty} \left(1 - \frac{x^2}{2m} \right)^{m} = e^{-x^2/2} $$ However, if $x = 2k\pi$ then the limit is $1$ since for all integer $n$ the value is $1$. And for $x = (2k+1)\pi$ the limit does not exist.

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  • $\begingroup$ What $n_0$ did you both use? :( Is it $n_0={2\over \pi}$? $\endgroup$ Jan 27 '15 at 22:35
  • $\begingroup$ There is no $n_0$ in the problem. The Taylor expansion is about $x_0 = 0$. $\endgroup$ Jan 27 '15 at 23:28
  • $\begingroup$ There is some formula to calculate those functions. As I understood, it is acceptable to look at the formula for $\cos x$ and then place, for example, ${c\over x}$, or to calculate it all with ${cos{c\over x}}$ to begin with? $\endgroup$ Jan 27 '15 at 23:34
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Noting that $\sin^2\left(\frac xn\right)=1-\cos^2\left(\frac xn\right)$ yields $$ \cos\left(\frac xn\right)=1-\frac{\sin^2\left(\frac xn\right)}{1+\cos\left(\frac xn\right)} $$ Therefore, $$ \begin{align} \left[\cos\left(\frac xn\right)\right]^{n^2} &=\left[1-\frac{n^2\sin^2\left(\frac xn\right)}{1+\cos\left(\frac xn\right)}\frac1{n^2}\right]^{n^2}\\ &=\left[1-\frac{a_n}{n^2}\right]^{n^2} \end{align} $$ where $a_n=\dfrac{n^2\sin^2\left(\frac xn\right)}{1+\cos\left(\frac xn\right)}$ and $\lim\limits_{n\to\infty}a_n=\dfrac{x^2}2$.

Since $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$ and the convergence is uniform on compact sets, we get that $$ \begin{align} \lim_{n\to\infty}\left[\cos\left(\frac xn\right)\right]^{n^2} &=\lim_{n\to\infty}\left[1-\frac{a_n}{n^2}\right]^{n^2}\\ &=\lim_{n\to\infty}\left[1-\frac{x^2}{2n^2}\right]^{n^2}\\[9pt] &=e^{-x^2/2} \end{align} $$

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If $x=0$ there's nothing to prove: the limit is $1$. Exchanging $x$ with $-x$ gives the same sequence, so we can assume $x>0$.

It's generally better to compute the limit of the logarithm of the sequence: $$ \lim_{n\to\infty}n^2\log\cos\frac{x}{n}. $$ If the limit $$ \lim_{y\to\infty} y^2\log\cos\frac{x}{y} $$ exists, then also our limit on the natural numbers exists and they're equal. Now we can do the substitution $t=x/y$, so the limit becomes $$ \lim_{t\to0^+}x^2\frac{\log\cos t}{t^2}. $$ Leaving aside the constant $x^2$, we can apply l'Hôpital's theorem: $$ \lim_{t\to0^+}\frac{\log\cos t}{t^2}= \lim_{t\to0^+}\frac{-\sin t/\cos t}{2t}= \lim_{t\to0^+}-\frac{\sin t}{t}\frac{1}{2\cos t}=-\frac{1}{2}. $$ Thus $$ \lim_{n\to\infty}n^2\log\cos\frac{x}{n}=-\frac{x^2}{2} $$ and the original limit is $e^{-x^2/2}$ (which also holds for $x\le0$ as we saw at the beginning).

Of course one can also do $$ \lim_{t\to0^+}\frac{\log\cos t}{t^2}= \lim_{t\to0^+}\frac{\log(1-t^2/2+o(t^2))}{t^2}= \lim_{t\to0^+}\frac{-t^2/2+o(t^2)}{t^2}=-\frac{1}{2}. $$

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