Let f(x) = * 20 c (1 + x) ^ 1 x - e x ;x 0 k , , , , , , , , , , … - Vidyadip

MCQ

Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{{(1 + \tan x)}^{\frac{1}{x}}} - e}}{x};x \ne 0}\\
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;x \ne 0}
\end{array}} \right.$ be continuous at $x = 0$, then $k$ is-

✓
$-\frac{e}{2}$
B
$-e$
C
$-\frac{e}{4}$
D
$\frac{e}{4}$

✓

Answer

Correct option: A.

$-\frac{e}{2}$

a
${\rm{k}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{{{\rm{e}}^{\frac{1}{{\rm{x}}}\ln (1 + \tan {\rm{x}})}} - {\rm{e}}}}{{\rm{x}}}} \right)$

${\rm{k}} = e\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{{{\rm{e}}^{\frac{1}{{\rm{x}}}\ln (1 + \tan {\rm{x}}) - 1}} - 1}}{{\rm{x}}}} \right)$

${\rm{k}} = e\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{{{\rm{e}}^{\frac{1}{{\rm{x}}}\ln (1 + \tan x) - 1}} - 1}}{{\frac{1}{{\rm{x}}}\ln (1 + \tan {\rm{x}}) - 1}}} \right)\left( {\frac{{\ln (1 + \tan {\rm{x}}) - {\rm{x}}}}{{{{\rm{x}}^2}}}} \right)$

$k=e \times\left(-\frac{1}{2}\right)=-\frac{e}{2}$

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