Compute the limit $\lim_{n\to\infty } n\left [\int_0^{\frac {\pi}{4}}\tan^n \left ( \frac{x}{n} \right )\mathrm {d}x\right]^{\frac{1}{n}}$ $$\lim_{n\to\infty } n\left [\int_0^{\frac {\pi}{4}}\tan^n  \left ( \frac{x}{n} \right )\mathrm {d}x\right]^{\frac{1}{n}}$$
 I want  to use squeeze  theorem to solve this problem, but I don't  know how to do  it.
 A: Hint. For $0 \leq x \leq \delta < \pi/2$, we have
$$ x \leq \tan x \leq \frac{\tan \delta}{\delta} x. $$
This easily follows from the convexity of $\tan x$ on $[0,\pi/2)$:
$\hspace{12em}$
Utilizing this inequality with $\delta = \frac{\pi}{4n}$, we can show that
$$ \frac{\pi}{4} \left( \frac{\pi}{4(n+1)} \right)^{\frac{1}{n}} \leq n \left[ \int_{0}^{\pi/4} \tan^n \left(\frac{x}{n}\right) \, \mathrm{d}x \right]^{\frac{1}{n}} \leq n \tan \left(\frac{\pi}{4n}\right) \left( \frac{\pi}{4(n+1)} \right)^{\frac{1}{n}}. $$
What happens if we take $n\to\infty$?
A: 
To use the squeeze theorem, we use the inequalities from THIS ANSWER for the tangent function
$$\bbox[5px,border:2px solid #C0A000]{x/n\le \tan(x/n)\le \frac{x/n}{\cos(x/n)} }\tag1$$
for $x/n\in [0,\pi/2]$.


Then using $(1)$, we have the upper bound estimates
$$\int_0^{\pi/4} \tan^n(x/n)\,dx\le \int_0^{\pi/4} \frac{(x/n)^n}{\cos^n(x/n)}\,dx\le \frac{1}{n^n}\frac{(\pi/4)^{n+1}}{(n+1)\cos^n(\pi/4n)}$$
from which we find
$$\bbox[5px,border:2px solid #C0A000]{n\left(\int_0^{\pi/4}\tan^n(x/n)\,dx\right)^{1/n}\le \frac{(\pi/4)^{1+1/n}}{(n+1)^{1/n}\cos(\pi/4n)}\to \pi/4}\tag 2$$

Similarly, we have the lower bound estimates
$$\int_0^{\pi/4} \tan^n(x/4)\,dx\ge \int_0^{\pi/4}(x/n)^n\,dx=\frac{1}{n^n}\frac{(\pi/4)^{n+1}}{(n+1)}$$
from which we find
$$\bbox[5px,border:2px solid #C0A000]{n\left(\int_0^{\pi/4}\tan^n(x/n)\,dx\right)^{1/n}\ge \frac{(\pi/4)^{1+1/n}}{(n+1)^{1/n}}\to \pi/4}\tag 3$$

Hence, using the squeeze theorem with $(2)$ and $(3)$, we find that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}n\left(\int_0^{\pi/4}\tan^n(x/n)\,dx\right)^{1/n}=\pi/4}$$

A: Here's a more general result, which has little to do with special properties of $\tan x.$ The point here is not to get fancy but to show what's really going on:
Thm: Suppose $f$ is continuous on $[0,b],$ with $f(0)=0, f'(0)=1.$ Then
$$\tag 1 \lim_{n\to \infty}n\left (\int_0^b f \left ( \frac{x}{n}\right)^n\,dx\right)^{1/n} = b.$$
Taking $f(x) = \tan x$ then shows the limit in the given problem is $\pi/4.$
Proof of theorem (sketch): Because $f(0)=0,f'(1)=1,$ it follows that $f(u)/u \to 1$ as $u\to 0.$ Another way to write this is $f(u)\sim u$ as $u\to 0.$ In the given expression, we are looking at $f(x/n),$ and $x/n$ is uniformly small on $[0,b]$ for large $n.$ It thus seems reasonable to replace $f(x/n)$ in the integral with $x/n.$ Doing that reduces the expression to 
$$n\left (\int_0^b (x/n)^n\, dx \right)^{1/n}  = \left (\int_0^b x^n\, dx\right)^{1/n}=\left (\frac{b^{n+1}}{n+1}\right)^{1/n}.$$
The limit of the last expression is $b.$
Well, that's not quite a proof; the word "reasonable" is a bit suspect. But it can be turned into a rigorous proof without too much trouble.
A: By the MVT for integrals, one has 
$$ \int_0^{\frac{\pi}{4}}\tan^n(\frac{x}{n})dx=\frac{\pi}{4}\tan^n(\frac{\xi_n}{n}),\xi_n\in(0,\frac{\pi}{4}). $$
So
$$  \int_0^{\frac{\pi}{4}}\tan^n(\frac{x}{n})dx=\frac{\pi}{4}\tan^n(\frac{\xi_n}{n})\le \frac{\pi}{4}\tan^n(\frac{\pi}{4n}).\tag{1} $$
On the other hand, noting that $\tan x\ge x$ for $x\in[0,\pi/2)$, one has
$$ \int_0^{\frac{\pi}{4}}\tan^n(\frac{x}{n})dx\ge\int_{0}^{\frac{\pi}{4}}(\frac{x}{n})^ndx=\frac{1}{n^n(n+1)}(\frac{\pi}{4})^{n+1}.\tag{2} $$
By (1)(2), one has
$$ \frac1{(n+1)^{\frac1n}}(\frac{\pi}{4})^{1+\frac1n}\le n\left(\int_0^{\frac{\pi}{4}}\tan^n(\frac{x}{n})dx\right)^{\frac{1}{n}}\le n(\frac{\pi}{4})^{\frac1n}\tan(\frac{\pi}{4n}). $$
Letting $n\to\infty$, one has
$$ \lim_{n\to\infty} n\left(\int_0^{\frac{\pi}{4}}\tan^n(\frac{x}{n})dx\right)^{\frac{1}{n}}=\frac{\pi}{4}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Laplace Method:

\begin{align}
&\lim_{n \to \infty}\braces{%
n\,\bracks{\int_{0}^{\pi/4}\tan^{n}\pars{x \over n}\,\dd x}^{1/n}}
\\[5mm] = &\
\lim_{n \to \infty}\braces{%
n\,\bracks{\int_{0}^{\pi/4}\exp\pars{n\ln\pars{\tan\pars{\pi/4 - x \over n}}}\, \dd x}^{1/n}}
\\[5mm] = &\
\lim_{n \to \infty}\pars{%
n\,\braces{\int_{0}^{\infty}\exp\pars{n\ln\pars{\tan\pars{\pi \over 4n}} -
\bracks{\cot\pars{\pi \over 4n} + \tan\pars{\pi \over 4n}}x}\, \dd x}^{\,1/n}}
\\[5mm] = &\
\lim_{n \to \infty}
{n\tan\pars{\pi/4n} \over
\bracks{\cot\pars{\pi/4n} + \tan\pars{\pi/4n}}^{\,1/n}} =
\bbox[15px,#efe]{\ds{\large{\pi \over 4}}}
\end{align}

Note that

$$
\left\{\begin{array}{rcl}
\ds{\lim_{n \to \infty}\bracks{n\tan\pars{\pi \over 4n}}} & \ds{=} &
\ds{\bbox[15px,#efe]{\ds{\large{\pi \over 4}}}}
\\[3mm]
\ds{\lim_{n \to \infty}\bracks{%
\cot\pars{\pi \over 4n} + \tan\pars{\pi \over 4n}}^{\,1/n}} & \ds{=} &
\ds{\lim_{n \to \infty}\pars{4n \over \pi}^{1/n} =
\bbox[15px,#efe]{\ds{\large 1}}}
\end{array}\right.
$$
A: According to the second integral mean value theorem, we have
$$
\int_0^{\frac{\pi}{4}}{\tan ^n\left( \frac{x}{n} \right)}\textrm{d}x=0\cdot \int_0^{\xi _n}{\textrm{d}x}+\tan ^n\left( \frac{\pi}{4n} \right) \int_{\xi _n}^{\frac{\pi}{4}}{\textrm{d}x}=\tan ^n\left( \frac{\pi}{4n} \right) \left( \frac{\pi}{4}-\xi _n \right) 
$$
Therefore, we have
$$
I=\lim_{n\rightarrow \infty}n\left[ \tan ^n\left( \frac{\pi}{4n} \right) \left( \frac{\pi}{4}-\xi _n \right) \right] ^{\frac{1}{n}}=\frac{\pi}{4}
$$
