Limit of a Function involving tangent function and limits at infinity

Question

Determine $$\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$.

Attempt

Let $$y=\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$

Put $\frac{1}{x}=p$.

$$\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.

We have $$\lim_{x \to \infty} y=\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.

Now consider the function $y$ in variable $p$ Taking $ln$ both sides $$ln\left(y\right)=p.ln\left(\tan{\frac{\pi}{2+p}}\right)$$. $$ln\left(y\right)=p.\frac{ln\left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}.\tan{\frac{\pi}{2+p}}$$.

Putting $\tan{\frac{\pi}{2+p}}=m$

We have

$$ln\left(y\right)=p.\frac{ln\left(m\right)}{m}.\tan{\frac{\pi}{2+p}}$$.

As $x \to \infty$ we have $p \to 0$ and hence $m \to \infty$

Hence the limit of $\frac{ln\left(m\right)}{m}$ is $0$.

But I am unable to show the limit of other to part of the product. Please help me out.

Answer 1

Note that

$$\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}=\left(\tan{\frac{\pi x+\frac{\pi}2-\frac{\pi}2}{2x+1}}\right)^\frac{1}{x}=\left(\tan{\left(\frac{\pi}2-\frac{\frac{\pi}2}{2x+1}\right)}\right)^\frac{1}{x}=\left(\tan{\frac{\pi}{4x+2}}\right)^{-\frac{1}{x}}=e^{-\frac{\log{\left(\tan{\frac{\pi}{4x+2}}\right)}}{x}}\to 1$$

Indeed

$$-\frac{\log{\left(\tan{\frac{\pi}{4x+2}}\right)}}{x}=-\left(\tan{\frac{\pi}{4x+2}}\right)\log{\left(\tan{\frac{\pi}{4x+2}}\right)}\frac{\frac{\pi}{4x+2}}{\tan{\frac{\pi}{4x+2}}}\frac{\frac{4x+2}{\pi}}{x}\to0\cdot1\cdot \frac4 \pi=0$$

As an alternative, according to the change of variable $p=\frac 1 x \to 0$, from here

$$\left(\tan{\frac{\pi}{2+p}}\right)^p=e^{p\log \left(\tan{\frac{\pi}{2+p}}\right)}\to 1$$

indeed

$$p\log \left(\tan{\frac{\pi}{2+p}}\right)=p\tan{\frac{\pi}{2+p}}\frac{\log \left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}\to 0$$

indeed

$$\frac{\log \left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}\to 0$$ $$p\tan{\frac{\pi}{2+p}}=\frac{p}{\tan\left({\frac{\pi}2-\frac{\pi}{2+p}}\right)}=\frac{p}{\tan\left({\frac{\pi p}{2p+4}}\right)}=\frac{\frac{\pi p}{2p+4}}{\tan\left({\frac{\pi p}{2p+4}}\right)}\frac{2p+4}{\pi}\to \frac 4 \pi$$

Answer 2

I would rather use asymptotic expansions.

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

As $x\to\infty$, $\frac{\pi x}{2x+1}\approx \pi/2$

For $h$ near to $\pi/2$, $$\sin(h)\approx 1$$ $$\cos(h)\approx \pi/2-h$$ $$\tan(h)\approx \frac1{\pi/2-h}$$

Thus, $$\tan( \frac{\pi x}{2x+1})\approx \frac1{\pi/2-\frac{\pi x}{2x+1}}=\frac2\pi(2x+1)$$

$$\ln y\approx \frac{\ln \frac2\pi(2x+1)}x\to 0$$

So, $y=1$.

Answer 3

Hint:

With $t:=1/x\to0$, $$\left(\tan\frac\pi{2+t}\right)^t=\left(\tan\left(\frac\pi2-\frac\pi {2+t}\right)\right)^{-t}=\left(\tan\frac{\pi t}{2(2+t)}\right)^{-t}$$

is asymptotic to $(\frac\pi4t)^{-t}$ and tends to $1$.

Answer 4

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

$$\lim_{x \to \infty}g(x) = \lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right) = \lim_{x \to \infty}\left(\frac{2 + 4 x}{\pi}\right)\tag{1}$$ So $$L = e^{\left(\lim_{x \to \infty}\frac{1}{x}\ln(g(x))\right)}$$ $$\lim_{x \to \infty}\left(\frac{\ln\left(\frac{2 + 4 x}{\pi}\right)}{x}\right) \stackrel{I´H}{\implies} \lim_{x \to \infty}\left(\frac{2}{2x + 1}\right) = 0$$ Therefore $$L = e^0 = 1$$

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(1) Proof :

For $x \to\infty$ $$g(x) = \tan\left(\frac{\pi x}{2x+1}\right) =\tan\left(\frac{\pi}{2}\frac{1 }{\left(1+\frac{1}{2x}\right)}\right) =\tan\left(\frac{\pi}{2}\frac{1 }{\left(1+\frac{1}{2x}\right) } - \frac{\pi}{2} + \frac{\pi}{2}\right)$$ $$\tan\left(\frac{\pi}{2}\left(\frac{1 }{\left(1+\frac{1}{2x}\right) } - 1\right) + \frac{\pi}{2}\right) = \tan\left(\frac{\pi}{2}- \frac{\pi}{2}\left(\frac{1}{2x+1 }\right)\right)$$

Set $\frac{1}{\left(2x+1\right) } = t$, observe that, as $t \to 0$
$$ \begin{align} \tan\left(\frac{\pi}{2} - \frac{\pi}{2}t\right) = \tan\left(\frac{\pi}{2}(1-t)\right) = \cot\left(\frac{\pi}{2}t\right) = \frac{\cos(\pi t) +1 }{\sin(\pi t)}=\\ \end{align}$$

$$\frac{\pi t}{\pi t}\cdot\frac{\cos(\pi t) +1 }{\sin(\pi t)}=$$ $$\frac{1}{t\pi}\cdot\underbrace{\left(\frac{\pi t }{\sin(\pi t)}\left(\cos(\pi t) +1\right)\right)}_{t \to 0 \implies 2}= \frac{1}{t\pi}\cdot 2 = $$ $$\frac{1}{\left(\frac{1}{\left(2x+1\right)}\right)\pi}\cdot 2 = \frac{(2 + 4 x)}{\pi}\tag*{$\Box$}$$