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I have to solve the limit with L'hospital's rule, but I can't. I understand that since there is a power here, I need to take a log from it.

Limit: $\lim _{x \rightarrow \infty} \sqrt[x]{\tan \left(\frac{\pi x}{2 x+1}\right)}$

After i taking a log, i get $e^{\lim _{x \rightarrow \infty} \frac{\ln \left(\tan \left(\frac{\pi x}{2 x+1}\right)\right)}{x}}$ which is $(\frac{\infty}{\infty})$, so i can use L'hospital's rule.

After I take derivatives and simplify them, I get such a limit:

$e^{\lim _{x \rightarrow \infty} \frac{\pi}{\cos \left(\frac{\pi x}{2 x+1}\right)(2 x+1)^{2}}}$

However, here I can no longer apply the L'hospital's rule anymore, but the indeterminate form hasn't disappeared since cos -> cos($\pi/2$) = 0, and $(2x+1)^2$ -> +inf.

I don't know how to proceed. Maybe I chose the wrong approach from the very beginning?

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I developed it as follows. Please, doublecheck if what I have done is fine.

\begin{eqnarray} \mathcal L &=& \lim_{x\to +\infty} \sqrt[x]{\tan\left(\frac{\pi x}{2x+1}\right)}=\\ &=&\lim_{x\to +\infty}\left[\tan \left(\frac{\pi}2-\frac{\pi}{4x+2}\right)\right]^{\frac1x}=\\ &=&\lim_{x\to+\infty}\left[\frac{\sin\left(\frac{\pi}2-\frac{\pi}{4x+2}\right)}{\cos\left(\frac{\pi}2-\frac{\pi}{4x+2}\right)}\right]^{\frac1x}=\\ &=&\lim_{x\to+\infty}\left[\frac{\cos\left(\frac{\pi}{4x+2}\right)}{\sin\left(\frac{\pi}{4x+2}\right)}\right]^{\frac1x}=\\ &=&e^{\lim_{x\to+\infty}\frac1x\log\left[\frac{\cos\left(\frac{\pi}{4x+2}\right)}{\sin\left(\frac{\pi}{4x+2}\right)}\right]}=\\ &=&e^{\lim_{x\to+\infty}\frac1x\log\left(\frac{4x+2}{\pi}\right)}=1, \end{eqnarray} where I used $$\cos\left(\frac{\pi}{4x+2}\right) \to 1$$ and $$\sin\left(\frac{\pi}{4x+2}\right)\sim \frac{\pi}{4x+2},$$ for $x\to+\infty$.

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  • $\begingroup$ Oh, thank you! I couldn't see the obvious step for so long :c $\endgroup$ – Alexander Nikolin Dec 16 '19 at 20:30
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Avoid writing complicated exponents and first do the substitution $x=1/t$, so the limit of the logarithm becomes $$ \lim_{t\to0^+}t\log\tan\dfrac{\pi t}{2+t} $$ For small enough $t$, this function is negative. Also, for small enough $\alpha$, $$ \tan\alpha>\alpha $$ so we have $$ t\log\frac{\pi t}{2+t}\le t\log\tan\dfrac{\pi t}{2+t}\le 0 $$ Now $$ t\log\frac{\pi t}{2+t}=t\log\pi+t\log t-t\log(2+t) $$ and $$ \lim_{t\to0^+}t\log\frac{\pi t}{2+t}= \lim_{t\to0^+}(t\log\pi+t\log t-t\log(2+t))=0 $$ By squeezing, also $$ \lim_{t\to0^+}t\log\tan\dfrac{\pi t}{2+t}=0 $$ and therefore your limit is $e^0=1$.

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  • $\begingroup$ But how do we know the tangent is greater for small values? from $\tan{a} \sim a$? $\endgroup$ – Alexander Nikolin Dec 17 '19 at 15:38
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    $\begingroup$ @AlexanderNikolin Let $g(x)=\tan x-x$; then $g(0)=0$ and $g'(x)=\tan^2x>0$ for $0<x<\pi/2$; thus the function $g$ is increasing and therefore positive for $0<x<\pi/2$. $\endgroup$ – egreg Dec 17 '19 at 15:42
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Or you could just use $\tan\frac{\pi x}{2x+1}=\cot\frac{\pi}{4x+2}\sim\frac{4x}{\pi}$ to prove the limit is $1$, viz.$$\lim_{x\to\infty}\frac{\ln(4x/\pi)}{x}=\lim_{x\to\infty}\frac{1}{x}=0.$$

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  • $\begingroup$ Thank you! I see that the main idea was to split the fraction from the beginning $\endgroup$ – Alexander Nikolin Dec 17 '19 at 8:07
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Write ´ $$e^{\lim_{x\to \infty}\frac{\ln\left(\tan\left(\frac{\pi x }{2x+1}\right)\right)}{x}}$$

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$$\lim_{x\to\infty}\left(\tan\dfrac{\pi x}{2x+1}\right)^{1/x}$$

$$=\left(\lim_{x\to\infty}\left(\tan\dfrac{\pi x}{2x+1}\right)^{\dfrac1{\tan\dfrac{\pi x}{2x+1}}}\right)^{\lim_{x\to\infty}\dfrac{\tan\dfrac{\pi x}{2x+1}}x}$$

For the inner limit, set $\tan\dfrac{\pi x}{2x+1}=n$

and use How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?

For the exponent, set $1/x=h$

$$\lim_{x\to\infty}\dfrac{\tan\dfrac{\pi x}{2x+1}}x=\lim_{h\to0^+}h\tan\dfrac{\pi}{2+h}=\lim_{h\to0^+}h\cot\dfrac{\pi h}{h+2}=\lim_{h\to0^+}\cos\dfrac{\pi h}{h+2}\cdot\lim_{h\to0^+}\dfrac h{\sin\dfrac{\pi h}{h+2}}=?$$

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