# Question 6637d

Feb 25, 2017

${\lim}_{x \rightarrow \frac{\pi}{4}} \frac{{e}^{\tan} x - e}{x - \frac{\pi}{4}} = 2 e$

#### Explanation:

I'm assuming you meant the limit:

${\lim}_{x \rightarrow \frac{\pi}{4}} \frac{{e}^{\tan} x - e}{x - \frac{\pi}{4}}$

Trying to evaluate at $x = \frac{\pi}{4}$ gives the indeterminate form $\frac{0}{0}$. Thus, we can solve this limit using L'Hopital's rule.

$= {\lim}_{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{\mathrm{dx}} \left({e}^{\tan} x - e\right)}{\frac{d}{\mathrm{dx}} \left(x - \frac{\pi}{4}\right)}$

$= {\lim}_{x \rightarrow \frac{\pi}{4}} \frac{{e}^{\tan} x \left({\sec}^{2} x\right)}{1}$

We can now evaluate the limit:

$= {e}^{\tan} \left(\frac{\pi}{4}\right) {\sec}^{2} \left(\frac{\pi}{4}\right)$

$= {e}^{1} {\left(\sqrt{2}\right)}^{2}$

$= 2 e$

Mar 2, 2017

$2 e .$

#### Explanation:

We will use the following Standard Limits to evaluate the reqd.

limit :

$\left(S L 1\right) : {\lim}_{h \to 0} \frac{{a}^{h} - 1}{h} = \ln a , a > 0.$

$\left(S L 2\right) : {\lim}_{\theta \to 0} \tan \frac{\theta}{\theta} = 1.$

Let, $l = {\lim}_{x \to \frac{\pi}{4}} \frac{{e}^{\tan} x - e}{x - \frac{\pi}{4}} .$

$= \lim \frac{e \left({e}^{\tan x - 1} - 1\right)}{x - \frac{\pi}{4}} ,$

$= \lim \left\{\frac{e \left({e}^{\tan x - 1} - 1\right)}{\tan x - 1}\right\} \left\{\frac{\tan x - 1}{x - \frac{\pi}{4}}\right\} ,$

$= e \left({l}_{1} {l}_{2}\right) , \ldots \ldots \ldots \left(\star\right)$ where,

${l}_{1} = {\lim}_{x \to \frac{\pi}{4}} \frac{{e}^{\tan x - 1} - 1}{\tan x - 1} .$

Subst. $h = \tan x - 1 , \text{ so that, as } x \to \frac{\pi}{4} , h \to 0.$

:. l_1=lim_(h to 0) (e^h-1)/h,

$= \ln e \ldots \ldots . . \left[\because , \left(S L 1\right)\right] ,$

$\therefore {l}_{1} = 1. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(\star 1\right) .$

For, ${l}_{2} , \text{ we take, "x=pi/4+theta," so; when,} x \to \frac{\pi}{4} , \theta \to 0.$

$\therefore {l}_{2} = {\lim}_{\theta \to 0} \frac{\tan \left(\frac{\pi}{4} + \theta\right) - 1}{\theta}$

$= \lim \frac{\frac{1 + \tan \theta}{1 - \tan \theta} - 1}{\theta}$

$= \lim \frac{2 \tan \theta}{\theta \left(1 - \tan \theta\right)}$

$= \lim \left(\tan \frac{\theta}{\theta}\right) \left(\frac{2}{1 - \tan \theta}\right)$

$= \left(1\right) \left(\frac{2}{1 - 0}\right) , \ldots \ldots \ldots \ldots \ldots \ldots \left[\because , \left(S L 2\right)\right]$

$\therefore {l}_{2} = 2. \ldots \ldots \ldots \ldots \ldots \ldots \left(\star 2\right) .$

$\text{From "(star1),(star2) & (star), "the reqd. lim. "l" follows as,}$

$l = 2 e .$

Enjoy Maths.!