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$$\lim_{n\to\infty } n\left [\int_0^{\frac {\pi}{4}}\tan^n \left ( \frac{x}{n} \right )\mathrm {d}x\right]^{\frac{1}{n}}$$ I want to use squeeze theorem to solve this problem, but I don't know how to do it.

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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 $[0,\pi/2)$:

$\hspace{12em}$enter image description here

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 $n\to\infty$?

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To use the squeeze theorem, we use the inequalities from THIS ANSWER for the tangent function

$$\bbox[5px,border:2px solid #C0A000]{x/n\le \tan^n(x/n)\le \frac{x/n}{\cos(x/n)} }\tag1$$

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

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

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Mar 21 '17 at 3:37
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Suppose $f$ is continuous and nonnegative on $[0,b],$ with $f(0)=0, f'(0)=1.$ Claim:

$$\tag 1 n\left (\int_0^b f \left ( \frac{x}{n}\right)^n\,dx\right)^{1/n} \to b.$$

Taking $f(x) = \tan x$ then shows the limit in our problem is $\pi/4.$

Proof of claim: Define

$$m_n= \inf_{y\in (0,b/n]}\frac{f(y)}{y}, \,\, M_n= \sup_{y\in (0,b/n]}\frac{f(y)}{y}.$$

Because $f(0)=0, f'(0)=1,$ both $m_n,M_n \to 1.$ We have the simple estimate $m_ny \le f(y) \le M_ny$ for $y\in [0,b/n].$

Now the change of variables $x=ny$ shows the left side of $(1)$ equals

$$ n^{(n+1)/n}\left(\int_0^{b/n}f(y)^n\,dy\right)^{1/n}.$$

Use the estimate $m_ny \le f(y) \le M_ny$ in the above, do a little work, and the claim falls right out using the squeeze theorem.

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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 $$ \lim_{n\to\infty} n\left(\int_0^{\frac{\pi}{4}}\tan^n(\frac{x}{n})dx\right)^{\frac{1}{n}}=\frac{\pi}{4}. $$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

$$ \left\{\begin{array}{rcl} \ds{\lim_{n \to \infty}\bracks{n\tan\pars{\pi \over 4n}}} & \ds{=} & \ds{\bbox[15px,#efe]{\ds{\large{\pi \over 4}}}} \\[3mm] \ds{\lim_{n \to \infty}\bracks{% \cot\pars{\pi \over 4n} + \tan\pars{\pi \over 4n}}^{\,1/n}} & \ds{=} & \ds{\lim_{n \to \infty}\pars{4n \over \pi}^{1/n} = \bbox[15px,#efe]{\ds{\large 1}}} \end{array}\right. $$

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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 $$ I=\lim_{n\rightarrow \infty}n\left[ \tan ^n\left( \frac{\pi}{4n} \right) \left( \frac{\pi}{4}-\xi _n \right) \right] ^{\frac{1}{n}}=\frac{\pi}{4} $$

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