# How do you solve lim_x->pi/4 (tanx-cotx)/(x-pi/4) ?

Jan 21, 2018

L'Hospital Rule yields:

${\lim}_{x \to \frac{\pi}{4}} \frac{\tan x - \cot x}{x - \frac{\pi}{4}} = 4$

#### Explanation:

by L'Hospital rule, which states that: when you get an indeterminate form after assuming that ${\lim}_{x \to c} f \left(x\right) = f \left(c\right)$, then ${\lim}_{x \to c} \frac{g \left(x\right)}{h \left(x\right)} = \frac{g ' \left(c\right)}{h ' \left(c\right)}$

$\implies {\lim}_{x \to \frac{\pi}{4}} \frac{{\left(\sec x\right)}^{2} + {\left(\csc x\right)}^{2}}{1}$

placing values of $x = \frac{\pi}{4}$

$= \frac{{\left(\sqrt{2}\right)}^{2} + {\left(\sqrt{2}\right)}^{2}}{1} = \frac{4}{1} = 4$

you can find the value of the limit as now it is not in the indertiminate form

hope you find it helpful :)

Aug 14, 2018

Here is a Solution without use of the L'Hospital's Rule.

#### Explanation:

Let, $x = \frac{\pi}{4} + y$.

$\therefore \text{ As } x \to \frac{\pi}{4} , y \to 0$.

Now, $\tan x - \cot x = \sin \frac{x}{\cos} x - \cos \frac{x}{\sin} x$,

$= \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x \cos x}$,

$= \frac{- 2 \left({\cos}^{2} x - {\sin}^{2} x\right)}{2 \sin x \cos x}$,

$= \frac{- 2 \cos 2 x}{\sin 2 x}$,

$= - 2 \cot 2 x$.

$\therefore \text{ The Reqd. Lim.} = {\lim}_{x \to \frac{\pi}{4}} \frac{- 2 \cot 2 x}{x - \frac{\pi}{4}}$,

$= {\lim}_{y \to 0} \frac{- 2 \cot \left(2 \left(\frac{\pi}{4} + y\right)\right)}{y}$,

$= {\lim}_{y \to 0} \frac{- 2 \cot \left(\frac{\pi}{2} + 2 y\right)}{y}$,

$= {\lim}_{y \to 0} \frac{- 2 \left(- \tan 2 y\right)}{y}$,

$= {\lim}_{y \to 0} \left\{2 \cdot \frac{\tan 2 y}{2 y} \cdot 2\right\}$.

Knowing that, ${\lim}_{\theta \to 0} \tan \frac{\theta}{\theta} = 1$,

$\text{The Lim.} = 4 \cdot 1 = 4$.

Aug 14, 2018

#### Explanation:

Solution without L'Hospital Rule :

Let,

$L = {\lim}_{x \to \frac{\pi}{4}} \frac{\tan x - \cot x}{x - \frac{\pi}{4}}$

Here,

$\tan x - \cot x = \sin \frac{x}{\cos} x - \cos \frac{x}{\sin} x = \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x \cos x}$

$\therefore \tan x - \cot x = - \frac{{\cos}^{2} x - {\sin}^{2} x}{2 \sin x \cos x} \times 2$

$\therefore \tan x - \cot x = - 2 \frac{\cos 2 x}{\sin 2 x} = - 2 \cot 2 x$

So,

$L = {\lim}_{x \to \frac{\pi}{4}} \frac{- 2 \cot 2 x}{\left(x - \frac{\pi}{4}\right)} = - 2 {\lim}_{x \to \frac{\pi}{4}} \frac{\cot 2 x}{\left(x - \frac{\pi}{4}\right)}$

Let , $x - \frac{\pi}{4} = \theta \implies x = \frac{\pi}{4} + \theta \implies 2 x = \frac{\pi}{2} + 2 \theta$

$\mathmr{and} x \to \frac{\pi}{4} \implies \theta \to 0$

$\therefore L = - 2 {\lim}_{\theta \to 0} \frac{\cot \left(\frac{\pi}{2} + 2 \theta\right)}{\theta} = - 2 {\lim}_{\theta \to 0} \frac{- \tan 2 \theta}{\theta}$

$\therefore L = 2 {\lim}_{\theta \to 0} \frac{\tan 2 \theta}{2 \theta} \times 2 = 2 \left(1\right) \times 2 = 4$