Evaluate the following limits: _ x 0 [ e^x+e^ -x -2 x x ]

Question

Evaluate the following limits:

$
\lim _{x \rightarrow 0}\left[\frac{e^x+e^{-x}-2}{x \cdot \tan x}\right]
$

✓

Answer

$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\mathrm{e}^x+\mathrm{e}^{-x}-2}{x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\mathrm{e}^x+\frac{1}{\mathrm{e}^x}-2}{x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\left(\mathrm{e}^x\right)^2+1-2\left(\mathrm{e}^x\right)}{\mathrm{e}^x \cdot x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\left(\mathrm{e}^x-1\right)^2}{\mathrm{e}^x \cdot x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\left[\frac{\left(\mathrm{e}^x-1\right)^2}{x^2}\right]}{\left(\frac{\mathrm{e}^x \cdot x \tan x}{x^2}\right)} \quad \ldots[\because x \rightarrow 0 ; x \neq 0] \\
= & \lim _{x \rightarrow 0} \frac{\left(\frac{\mathrm{e}^x-1}{x}\right)^2}{\mathrm{e}^x \cdot\left(\frac{\tan x}{x}\right)} \\
= & \frac{(1)^2}{\mathrm{e}^0 \cdot 1} \\
= & 1 \\
= & \left.\lim _{x \rightarrow 0} \mathrm{e}^x \cdot \lim _{x \rightarrow 0} \frac{\mathrm{e}^x-1}{x}\right)^2
\end{aligned}
$

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