Lim _ k 0 , ,1 k _0^k (1 + 2x) ^ 1 x dx - Vidyadip

MCQ

$\mathop {Lim}\limits_{k \to 0} \,\,\frac{1}{k}\int\limits_0^k {{{(1 + \sin 2x)}^{\frac{1}{x}}}dx} $

A
$2$
B
$1$
✓
$e^2$
D
non existent

✓

Answer

Correct option: C.

$e^2$

c
$l =$ $\mathop {Lim}\limits_{k \to 0} \,\,\frac{{\int\limits_0^k {{{(1 + \sin 2x)}^{\frac{1}{x}}}dx} }}{k}$
differentiating Using $ L’$ opital rule

$ l = $ $\mathop {Lim}\limits_{k\, \to \,0} \,{(1 + \sin 2k)^{\frac{1}{k}}}$ = ${e^{\mathop {Lim}\limits_{k \to 0} \,\frac{1}{k}(\sin 2k)}}$ = $e^2$

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