Find $\lim_{n\to \infty}(\cos{x\over n})^{n^2}$

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.

• What do you mean by "I tried using $e$"? The number $e$ is not a mathematical technique which can be used to solve problems. Jan 27 '15 at 21:48
• @JimmyK4542 He is probably referring to $x=e^{\log x}$? Jan 27 '15 at 21:52
• 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. Jan 27 '15 at 22:00

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 $n\to\infty$

• I can't make it :( I don't know what $x_0$ I should pick Jan 27 '15 at 22:05
• For $n\to\infty$, $x/n\to0$. So $x_0=0$, if I understood your question correctly. Jan 27 '15 at 22:11
• When I differentiate I got a really big expression. If I pick $n_0=0$ It is not defined... Jan 27 '15 at 22:20
• I accidentally used the whole expression oops :$Jan 27 '15 at 22:22 Combine your two ideas! This demonstration is not altogether rigorous but it does get the right answer: Assume$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. • What$n_0$did you both use? :( Is it$n_0={2\over \pi}$? Jan 27 '15 at 22:35 • There is no$n_0$in the problem. The Taylor expansion is about$x_0 = 0$. Jan 27 '15 at 23:28 • 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? Jan 27 '15 at 23:34 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} wherea_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^xand 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} Ifx=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 $$\lim_{t\to0^+}\frac{\log\cos t}{t^2}= \lim_{t\to0^+}\frac{\log(1-t^2/2+o(t^2))}{t^2}= \lim_{t\to0^+}\frac{-t^2/2+o(t^2)}{t^2}=-\frac{1}{2}.$$