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$\displaystyle f(x) = \lim_{n \to \infty} ((\cos x )^n + (\sin x)^n))^{1/n}$. Then $\int _0 ^{\pi /2} f(x)\,dx$ is?

I proceeded by attempting to solve the limit first as the integral for the function with the powers is very difficult to solve but even using l'hospital twice (after taking log) I am not able to resolve the limit. How do I proceed?

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  • $\begingroup$ $$\int^{\frac{\pi}{4}}_{0}()dx+\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}()dx$$ $\endgroup$ – jacky Feb 6 at 9:30
  • $\begingroup$ Sin and cos under those ranges are still greater than zero/less than one so the power $((\sin x)^n + (\cos x)^n)$ would still tend to 0. So I would end up solving the limit again? $\endgroup$ – user619072 Feb 6 at 9:35
  • $\begingroup$ If $x=0$ or $x=\frac{\pi}{2}$ the limit is $1$. For other values of $x$ you have an indeterminate limit of the type $0^0$. $\endgroup$ – PierreCarre Feb 6 at 9:44
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$$f(x)=\lim_{n\rightarrow \infty}\bigg((\cos x)^n+(\sin x)^n\bigg)^{\frac{1}{n}}$$

$$=\left\{\begin{matrix} \lim_{n\rightarrow \infty}\cos x\bigg[(\tan x)^n+1\bigg]^{\frac{1}{n}}=\cos x\;, \displaystyle x\in \bigg(0,\frac{\pi}{4}\bigg) & \\\ \lim_{n\rightarrow \infty}\sin x\bigg[(\cot x)^n+1\bigg]^{\frac{1}{n}}=\sin x\;, \displaystyle x\in \bigg(\frac{\pi}{4},\frac{\pi}{2}\bigg) & \end{matrix}\right.$$

$$\int^{\frac{\pi}{2}}_{0}f(x)dx = \int^{\frac{\pi}{4}}_{0}\cos xdx +\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\sin xdx$$

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$\lim (a^{n}+b^{n})^{1/n}=\max \{a,b\}$ for any $a , b>0$ which makes the question very easy to answer. To prove this take $a<b$ and write $(a^{n}+b^{n})^{1/n}=a(1+(b/a)^{n})^{1/n}$; use the fact that $\log(1+x)$ behaves like $x$ for $x$ small.

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    $\begingroup$ Kavi.Very nice answer. Could use $a \lt a(1+(b/a)^b)^{1/n} \lt a(2)^{1/n}.$ $\endgroup$ – Peter Szilas Feb 6 at 9:55
  • $\begingroup$ @PeterSzilas Agreed. Thanks. $\endgroup$ – Kavi Rama Murthy Feb 6 at 9:57
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For numbers $a,b \ge 0$ we have

$(a^n+b^n)^{1/n} \to \max \{a,b\}$ as $n \to \infty.$. Hence

$f(x)= \max \{ \cos x, \sin x\}$ for $x \in [0, \pi/2].$ This gives

$f(x)= \cos x$ $x \in [0, \pi/4]$ and $f(x)= \sin x$ for $ x \in [\pi/4,\pi/2].$

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  • $\begingroup$ Check the last line, $[\pi/4,\pi/2]$ maybe? $\endgroup$ – Oscar Lanzi Feb 6 at 10:35
  • $\begingroup$ Ooops ! Thanks ! $\endgroup$ – Fred Feb 6 at 11:47

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