Question

\[\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{\tan x-x}=\]

Options

• 1

• 0

• \[\frac{1}{2}\]

• \[\frac{1}{4}\]

Answer 1

1

Explanation:

= \[\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{\tan x-x}\]

\[=\lim_{x\to0}\frac{\mathrm{e}^{x}\left(\mathrm{e}^{\tan x-x}-1\right)}{\tan x-x}\]

\[=\mathrm{e}^0\cdot1\quad\ldots \begin{bmatrix} \operatorname{as}x\to0,\operatorname{tan}x-x\to0 \\ \\ \lim_{x\to0}\frac{\mathrm{e}^x-1}{x}=1 \end{bmatrix}\]