Question

Evaluate the following :

${\displaystyle \underset{x\to 0}{lim}\left[\frac{{\text{e}}^x+{\text{e}}^{-x}-2}{x\cdot \tan x}\right]}$

Answer 1

${\displaystyle \underset{x\to 0}{lim}\frac{{\text{e}}^x+{\text{e}}^{-x}-2}{x\tan x}}$

= ${\displaystyle \underset{x\to 0}{lim}\frac{{\text{e}}^x+\frac{1}{{\text{e}}^x}-2}{x\tan x}}$

= ${\displaystyle \underset{x\to 0}{lim}\frac{{\left({\text{e}}^x\right)}^2+1-2\left({\text{e}}^x\right)}{{\text{e}}^x\cdot x\tan x}}$

= ${\displaystyle \underset{x\to 0}{lim}\frac{{({\text{e}}^x-1)}^2}{{\text{e}}^x\cdot x\tan x}}$

= ${\displaystyle \underset{x\to 0}{lim}\frac{\left(\frac{{({\text{e}}^x-1)}^2}{x^2}\right)}{\left(\frac{{\text{e}}^x\cdot x\tan x}{x^2}\right)}}$  ...[∵ x → 0; ∴ x ≠ 0]

= ${\displaystyle \underset{x\to 0}{lim}\frac{{\left(\frac{{\text{e}}^x-1}{x}\right)}^2}{{\text{e}}^x\cdot \left(\frac{\tan x}{x}\right)}}$

= ${\displaystyle \frac{{\left(\underset{x\to 0}{lim}\frac{{\text{e}}^x-1}{x}\right)}^2}{\underset{x\to 0}{lim}{\text{e}}^x\cdot \underset{x\to 0}{lim}\left(\frac{\tan x}{x}\right)}}$

= ${\displaystyle \frac{{\left(1\right)}^2}{{\text{e}}^0\cdot 1}}$

= 1