The value of _ x 0 2 x (1 + x) is equal to - Vidyadip

MCQ

The value of $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x)$ is equal to

A
$e$
B
${e^2}$
C
$\frac{1}{2}$
✓
$2$

✓

Answer

Correct option: D.

$2$

d
(d) $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x) = \mathop {\lim }\limits_{x \to 0} 2\log {(1 + x)^{\frac{1}{x}}}$

$ = \mathop {\lim }\limits_{x \to 0} 2{\log _e}e = 2$

$\left\{ { \because \mathop {\lim }\limits_{x \to 0} {{(1 + x)}^{\frac{1}{x}}} = {{\log }_e}e = 1} \right\}$

Trick : Using $L$ Hospital’s rule.

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