Why does $e^{-(x^2/2)} \approx \cos[\frac{x}{\sqrt{n}}]^n$ hold for large $n$?

Question

Why does this hold: $$ e^{-x^2/2} = \lim_{n \to \infty} \cos^n \left( \frac{x}{\sqrt{n}} \right) $$

I am not sure how to solve this using the limit theorem.

Answer 1

$\cos(x)$ is the characteristic function of a signed Bernoulli.

Answer 2

In a neighbourhood of the origin, $$\log\cos z = -\frac{z^2}{2}\left(1+O(z^2)\right)\tag{1} $$ hence for any $x$ and for any $n$ big enough: $$ \log\left(\cos^n\frac{x}{\sqrt{n}}\right)=-\frac{x^2}{2}\left(1+O\left(\frac{1}{n}\right)\right)\tag{2}$$ and the claim follows by exponentiating $(2)$: $$ \cos^n\frac{x}{\sqrt{n}} = e^{-x^2/2}\cdot\left(1+O\left(\frac{1}{n}\right)\right).\tag{3}$$

Answer 3

Not as elegant as Jack's answer, but rewrite the limit as

$$\lim_{n\rightarrow \infty} e^{\dfrac{\ln(\cos(x/\sqrt{n}))}{1/n}}$$

and use the continuity of $e^x$ to perform l'Hospital's rule twice.

Answer 4

Hint: Use the Taylor series approximation for $~\cos t\simeq1-\dfrac{t^2}2~$ when $t\to0$, in conjunction with the limit definition of $~e=\displaystyle\lim_{u\to\infty}\bigg(1+\frac1u\bigg)^u$

Answer 5

$$ \begin{align} \cos^n\!\left(\frac{x}{\sqrt{n}}\right) &=\left(1-\sin^2\left(\frac{x}{\sqrt{n}}\right)\right)^{n/2}\\ &=\left(1-\frac{x^2}n\color{#C00000}{\left[\frac{\sin\left(\frac{x}{\sqrt{n}}\right)}{\frac{x}{\sqrt{n}}}\right]^2}\right)^{n/2}\\ \end{align} $$ Since $$ \lim_{n\to\infty}\frac{\sin\left(\frac{x}{\sqrt{n}}\right)}{\frac{x}{\sqrt{n}}}=1 $$ we can choose an $n$ large enough that the expression in red is as close to $1$ as we wish. Therefore, because $e^x$ is continuous for all $x$, we get $$ \begin{align} \lim_{n\to\infty}\cos^n\!\left(\frac{x}{\sqrt{n}}\right) &=\lim_{n\to\infty}\left(1-\frac{x^2}n\right)^{n/2}\\[6pt] &=e^{-x^2/2} \end{align} $$

Answer 6

Recall that if $t\in[0,\frac\pi2)$ then $$ \sin t \le t \le \tan t $$ Apply $\int_0^x \cdot \,dt$ to obtain that if $x\in[0,\frac\pi2)$ then $$ 1 - \cos x \le \frac{x^2}{2} \le \ln\sec x $$ Rearranging yields that if $x\in[0,\frac\pi2)$ then $$ 1 - \frac{x^2}{2} \le \cos x \le e^{-x^2/2} $$ Everything is even, so in fact this holds for $x\in(-\frac\pi2,\frac\pi2)$. For general $x$, we have $\frac{x}{\sqrt n}\in (-\frac\pi2,\frac\pi2)$ for all sufficiently large $n$, and so $$ 1 - \frac{x^2}{2n} \le \cos\Big(\frac{x}{\sqrt n}\Big) \le e^{-x^2/2n} $$ for sufficiently large $n$. Raising both sides to the power $n$ and applying the squeeze theorem finishes the job.

Answer 7

For sure, this is not as elegant as previous answers but I love too much Taylor expansions !

Starting with $$ \cos \left( \frac{x}{\sqrt{n}} \right)=1-\frac{x^2}{2 n}+\frac{x^4}{24 n^2}-\frac{x^6}{720 n^3}+\frac{x^8}{40320 n^4}-\frac{x^{10}}{3628800 n^5}+O\left(x^{12}\right)$$ and raising to power $n$ (using binomial theorem) $$ \cos^n \left( \frac{x}{\sqrt{n}} \right)=1+\sum_{n=1}^\infty a_n x^{2n}$$ with $$a_1=-\frac 12$$ $$a_2=\frac{3n-2}{24n}\to \frac 18$$ $$a_3=-\frac{15 n^2-30 n+16}{720 n^2}\to -\frac 1{48} $$ $$a_4=\frac{105 n^3-420 n^2+588 n-272}{40320 n^3}\to \frac 1{384}$$ $$a_5=-\frac{945 n^4-6300 n^3+16380 n^2-18960 n+7936}{3628800 n^4}\to -\frac 1{3840}$$ $$a_6=\frac{10395 n^5-103950 n^4+429660 n^3-893640 n^2+911328 n-353792}{479001600 n^5}\to \frac 1{46080}$$