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When applying L'Hopital's rule first I get this: $\lim_{n\to\infty}\log f_n(x)=\lim_{n\to\infty}-\frac {\sin\big (\frac x {\sqrt n} \big ) x\sqrt n}{2\cos \big ( \frac x {\sqrt n}\big )} $

by applying it second time I get;

$\lim_{n\to\infty} \log f_n(x)=\lim_{n\to\infty}\big (\frac {x\cos \big ( \frac x {\sqrt n} \big ) \sqrt n } {2\sin \big ( \frac x {\sqrt n} \big ) }-\frac n 2 \big )$

So now from here how to get:

$\lim_{n\to\infty}f_n(x)=e ^{-\frac {x^2} 2}$

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  • $\begingroup$ making a change of variables $t=\frac{1}{\sqrt{n}}$ may simplify the calculation and then apply L'Hospital rule (may be twice). Result is $x^2/2$. $\endgroup$ – daulomb Dec 4 '17 at 1:31
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    $\begingroup$ @daulomb: I got $-x^2/2$ (using your substitution). The minus sign comes from the fact that the derivative of $\ln{\left(\cos{x}\right)}$ is $-\tan{x}$. $\endgroup$ – Ant Dec 4 '17 at 1:40
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    $\begingroup$ yes you are right it comes from the derivative of cosine $\endgroup$ – daulomb Dec 4 '17 at 1:41
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We have $f_n(x)={\left(\cos{\left(\frac{x}{\sqrt{n}}\right)}\right)}^n$. Thus, $\ln{f_n(x)}=n\ln{\left(\cos{\left(\frac{x}{\sqrt{n}}\right)}\right)}$ and:

$$\ln{L(x)}:=\ln{\left(\lim_{n\to\infty}f_n(x)\right)}=\lim_{n\to\infty}{n\ln{\left(\cos{\left(\frac{x}{\sqrt{n}}\right)}\right)}}$$

Using the substitution suggested by daulomb, we have $t=\frac{1}{\sqrt{n}}$, so as $n$ approaches $\infty$, $t$ approaches $0$. Thus:

$$\ln{L(x)}=\lim_{t\to0}{\frac{1}{t^2}\ln{\left(\cos{\left(tx\right)}\right)}}$$

Now we use L'Hopital's rule, since we have a $0/0$ indeterminate form (remember that we take the derivative with respect to $t$, not $x$):

$$\ln{L(x)}=\lim_{t\to0}{\frac{\frac{-x\sin{\left(tx\right)}}{\cos{\left(tx\right)}}}{2t}}=-x\lim_{t\to0}{\frac{\tan{\left(tx\right)}}{2t}}$$

We took the $-x$ outside of the limit because it doesn't depend on $t$. Now we apply L'Hopital's rule again:

$$\ln{L(x)}=-x\lim_{t\to0}{\frac{x\sec^2{\left(tx\right)}}{2}}=-\frac{x^2}{2}\lim_{t\to0}{\sec^2{\left(tx\right)}}=-\frac{x^2}{2}$$

thus $L(x)=e^{-\frac{x^2}{2}}$.

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Alternative method. Without applying L'Hopital, but instead Taylor expansions (around $0$) of standard functions: $$ \begin{align} \cos u &= 1-\frac{u^2}{2} + o(u^2)\\ \ln(1+u) &= u +o(u)\end{align} $$ we get, as $\lim_{n\to\infty}\frac{x}{\sqrt{n}} = 0$ (for any fixed $x$): $$ \ln\cos\frac{x}{\sqrt{n}} = \ln\left(1-\frac{x^2}{2n} + o\left(\frac{1}{n}\right)\right)\tag{1}$$ and from there, since $\lim_{n\to\infty}\frac{x^2}{2n} = 0$ $$ \ln\cos\frac{x}{\sqrt{n}} = -\frac{x^2}{2n} + o\left(\frac{1}{n}\right)\,.\tag{2} $$ Therefore, for any fixed $x\in\mathbb{R}$, $$ \ln f_n(x)=\frac{\ln\cos\frac{x}{\sqrt{n}}}{\frac{1}{n}} = \frac{-\frac{x^2}{2n} + o\left(\frac{1}{n}\right)}{\frac{1}{n}} = -\frac{x^2}{2} + o\left(1\right) \xrightarrow[n\to\infty]{} \boxed{-\frac{x^2}{2}}. $$ By continuity of $\exp$, this immediately implies that $$ \boxed{\lim_{n\to\infty}f_n(x) = e^{-\frac{x^2}{2}}} $$

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  • $\begingroup$ as in here, from which this question was derived ;) +1 math.stackexchange.com/questions/2549673/… $\endgroup$ – qbert Dec 4 '17 at 1:50
  • $\begingroup$ @qbert Oh damn. I definitely missed that. $\endgroup$ – Clement C. Dec 4 '17 at 1:51
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    $\begingroup$ it's ok, I think it is the more natural thing to do here $\endgroup$ – qbert Dec 4 '17 at 1:51

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