A calculus problem by Hamza A

Let $L=\displaystyle\lim_{n\rightarrow \infty} {\sqrt[n]{ (sin x)^{n}+(cos x)^{n}}}$

$L=\displaystyle\lim_{n\rightarrow \infty}{cos x}{\sqrt[n]{ 1+(tan x)^{n}}}$

Now let $L_{1}=\displaystyle\lim_{n\rightarrow \infty} {\sqrt[n]{ 1+(tan x)^{n}}}$

Taking $ln$ both sides

$lnL_{1}=\large\displaystyle\lim_{n\rightarrow \infty}\displaystyle\dfrac{ln(1+ \tan^{n}{x})}{n}$

for $0<x<π/4 , 0<tan x<1, lnL_{1}$ goes to 0 hence $L_{1}$ is 1 and $L$ is $\large cos x$ .

For $π/4<x<π/2 , tan x>1 , lnL_{1}$ goes to $lntan x$ using L'hopital rule hence $L$ is $cos x tan x$ i.e. $L=sin x$ .

rest work is easy

We get $\large\displaystyle\int_{0}^{π/4}$ $cos x dx$ + $\large\displaystyle\int_{π/4}^{π/2}$ $sin x dx$

$=1/√2+1/√2$

$=√2$

Let $\displaystyle I = \int_{0}^{\frac{\pi}{2}} (\sin^n(x)+\cos^n(x))^\frac{1}{n} dx$ .

Let us substitute $u = \frac{\pi}{2} - x$ . (We find that $dx = -du$ .)

$\begin{aligned} I &= \int_{0}^{\frac{\pi}{2}} (\sin^n(x)+\cos^n(x))^\frac{1}{n} dx \\ &= \int_{\frac{\pi}{2}}^{0} \Big(\sin^n\Big(\frac{\pi}{2} - u\Big) + \cos^n\Big(\frac{\pi}{2} - u\Big)\Big)^\frac{1}{n} (-du) \\ &= \int_{0}^{\frac{\pi}{2}} (\cos^n(u)+\sin^n(u))^\frac{1}{n} du \\ &= I \end{aligned}$

We can think of the substitution $u = \frac{\pi}{2} - x$ as reflecting the graph of $\displaystyle (\sin^n(x)+\cos^n(x))^\frac{1}{n}$ in the line $x=\frac{\pi}{4}$ . We see that integral does not change. This is because the graph of $\displaystyle (\sin^n(x)+\cos^n(x))^\frac{1}{n}$ is symmetric along the line $x=\frac{\pi}{4}$ .

Hence, we deduce that $\displaystyle I = 2\int_{0}^{\frac{\pi}{4}} (\sin^n(x)+\cos^n(x))^\frac{1}{n} dx$ .

Now, we have to figure out what happens as $n$ tends to infinity.

$\begin{aligned} I &= 2\int_{0}^{\frac{\pi}{4}} (\sin^n(x)+\cos^n(x))^\frac{1}{n} dx \\ &= 2\int_{0}^{\frac{\pi}{4}} \cos(x)(\tan^n(x)+1)^\frac{1}{n} dx \\ &= 2\int_{0}^{\frac{\pi}{4}} \cos(x)\exp\bigg(\frac{\ln(1+\tan^n(x))}{n}\bigg) dx \end{aligned}$

We observe the following chain of implications.

$\begin{aligned} 0 \leq x \leq \frac{\pi}{4} & \implies 0 \leq \tan(x) \leq 1 \\ & \implies 0 \leq \tan^n(x) \leq 1 & \text{where } n \text{ is a positive real number} \\ & \implies 1 \leq \tan^n(x) + 1 \leq 2 \\ & \implies 0 \leq \ln(\tan^n(x) + 1) \leq \ln(2) & \text{since } \ln(x) \text{ is a strictly increasing function} \\ & \implies 0 \leq \frac{\ln(1+\tan^n(x))}{n} \leq \frac{\ln(2)}{n} \\ \end{aligned}$

We know that $\displaystyle \lim_{n\to\infty} 0 = 0$ and $\displaystyle \lim_{n\to\infty} \frac{\ln(2)}{n} = 0$ . Therefore, by the squeeze theorem

$\lim_{n\to\infty} \frac{\ln(1+\tan^n(x))}{n} = 0$

Consequently,

$\begin{aligned} \bigg(\lim_{n\to\infty} 2\int_{0}^{\frac{\pi}{4}} \cos(x)\exp\bigg(\frac{\ln(1+\tan^n(x))}{n}\bigg) dx\bigg)^2 &= \bigg(2\int_{0}^{\frac{\pi}{4}} \cos(x)\exp\bigg(\lim_{n\to\infty} \frac{\ln(1+\tan^n(x))}{n} \bigg) dx \bigg)^2 \\ & = \bigg(2\int_{0}^{\frac{\pi}{4}} \cos(x) dx\bigg)^2 \\ &= \Big(2\Big(\frac{1}{\sqrt{2}}\Big)\Big)^2 \\ &= \boxed{2} \end{aligned}$

Take out (cosx)^n from (sinx^n+cosx^n)^(1/n) and break limits at x=π/4 . For x<π/4 , tanx^n<<<<<<.........1 ,(tanx is a proper fraction and n tends to infinity) and for x>π/4 , tanx^n>>>>>>>>>>>>>>>.......1 (tanx is greater than 1 and n tends to infinity). The integral has reduced to : Integral of (cosx from 0 to π/4) + integral of (sinx from π/4 to π/2) = √2