# Trigonometric limit with L'hopital's rule and indeterminate form $\infty^{0}$

I have to solve the limit with L'hopital'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'hopital'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?

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 $$x\to+\infty$$.

• Oh, thank you! I couldn't see the obvious step for so long :c Commented Dec 16, 2019 at 20:30

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 $$e^0=1$$.

• But how do we know the tangent is greater for small values? from $\tan{a} \sim a$? Commented Dec 17, 2019 at 15:38
• @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 $0<x<\pi/2$. Commented Dec 17, 2019 at 15:42

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.$$\lim_{x\to\infty}\frac{\ln(4x/\pi)}{x}=\lim_{x\to\infty}\frac{1}{x}=0.$$

• Thank you! I see that the main idea was to split the fraction from the beginning Commented Dec 17, 2019 at 8:07

Write ´ $$e^{\lim_{x\to \infty}\frac{\ln\left(\tan\left(\frac{\pi x }{2x+1}\right)\right)}{x}}$$

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

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

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

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

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