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Prove that $\lim_{x \to \frac{\pi}2}\left(\tan x - \frac{4\pi}{\pi^2-4x^2}\right) = - \frac{1}\pi$

It is $\infty -\infty$ form, so I try to make $0/0$ form then try to use LH rule:

$$\lim_{x \to \frac{\pi}2}\frac{\pi^2\tan x-4x^2 \tan x-4\pi}{\pi^2-4x^2} \\ =\lim_{x \to \frac{\pi}2}\frac{\pi^2\sec^2 x -8x\tan x -4x^2 \sec^2x}{-8x} $$

I am not sure is the first line is $0/0$ form or not . I apply LH Rule, everything messed up. Wolfrom alpha conform the limit exist. I want to ask why simplifying the limit make it limit doesn't exist situation? Thanks for your time.

O M
  • 2,274

2 Answers2

3

$\begin{gathered} L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \left( {\tan \left( x \right) - \frac{{4\pi }}{{{\pi ^2} - 4{x^2}}}} \right),{\text{ let}}:y = x - \frac{\pi }{2},x \to \frac{\pi }{2} \Rightarrow y \to 0 \hfill \\ L = \mathop {\lim }\limits_{y \to 0} \left( { - \cot \left( y \right) + \frac{\pi }{{y\left( {y + \pi } \right)}}} \right) = \mathop {\lim }\limits_{y \to 0} \left( {\frac{{\pi - y\left( {y + \pi } \right)\cot \left( y \right)}}{{y\left( {y + \pi } \right)}}} \right)\left( {\frac{0}{0}} \right) \hfill \\ \overbrace = ^{{\text{L'H}}}\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\frac{{{y^2}}}{{{{\sin }^2}\left( y \right)}} - 2\cos \left( y \right)\frac{y}{{\sin \left( y \right)}} - \pi \left( {\frac{{\cos \left( y \right)}}{{\sin \left( y \right)}} - \frac{y}{{{{\sin }^2}\left( y \right)}}} \right)}}{{\pi + 2y}}} \right) \hfill \\ \mathop {\lim }\limits_{y \to 0} \frac{y}{{\sin \left( y \right)}} = 1 \Rightarrow \mathop {\lim }\limits_{y \to 0} \frac{{{y^2}}}{{{{\sin }^2}\left( y \right)}} = 1{\text{ and }}\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\cos \left( y \right)}}{{\sin \left( y \right)}} - \frac{y}{{{{\sin }^2}\left( y \right)}}} \right) = \mathop {\lim }\limits_{y \to 0} \frac{{\sin \left( y \right)\cos \left( y \right) - y}}{{{{\sin }^2}\left( y \right)}} \hfill \\ = 4\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\sin \left( {2y} \right) - 2y}}{{{{\left( {2y} \right)}^3}}}\frac{{{y^2}}}{{{{\sin }^2}\left( y \right)}}y} \right) = 0\because \mathop {\lim }\limits_{y \to 0} \frac{{\sin \left( y \right) - y}}{{{y^3}}} = - \frac{1}{6}{\text{ is known}} \hfill \\ \Rightarrow L = \frac{{1 - 2 - \pi .0}}{{\pi + 2.0}} = - \frac{1}{\pi } \hfill \\ \end{gathered} $

OnTheWay
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2

Let $x=\frac \pi 2-y$ with $y\to 0$ so that

$$\lim_{x \to \frac{\pi}2}\left(\tan x - \frac{4\pi}{\pi^2-4x^2}\right)=\lim_{y \to 0}\left(\frac1{\tan y} - \frac{\pi}{y(\pi-y)}\right)$$

and by standard limits

$$\frac1{\tan y} - \frac{\pi}{y(\pi-y)}=\frac{\pi(y-\tan y)}{y\tan y(\pi-y)} -\frac{y}{\tan y(\pi-y)}\to 0-\frac1\pi =-\frac1\pi$$

since

$$\frac{\pi(y-\tan y)}{y\tan y(\pi-y)}=\frac{\pi}{\pi-y}\frac{y-\tan y}{y^3}\frac{y^3}{y\tan y}\to 1\cdot\left(-\frac13\right)\cdot 0= 0$$

user
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