MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = $
- ✓$1$
- B$e$
- C${e^{ - 1}}$
- D$0$
✓
Answer
Correct option: A.
$1$
a
(a) $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^x}[{e^{\tan x - x}} - 1]}}{{\tan x - x}}$
(a) $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^x}[{e^{\tan x - x}} - 1]}}{{\tan x - x}}$
$ = \mathop {\lim }\limits_{x \to 0} \,{e^x}\,.\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\tan x - x}} - 1}}{{\tan x - x}} = {e^0} \times 1 = 1$.
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