_ x 0 x e^x - (1 + x) x^2 equals - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$ equals

A
$\frac{2}{3}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
✓
$\frac{3}{2}$

✓

Answer

Correct option: D.

$\frac{3}{2}$

d
(d) Let $y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{x\,{e^x} - \log \,(1 + x)}}{{{x^2}}}$, $\left( {\frac{0}{0}\,{\rm{form}}} \right)$

Applying $ L-$ Hospital's rule,

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^x} + x\,{e^x} - \frac{1}{{1 + x}}}}{{2x}}$, $\left( {\frac{0}{0}\,{\rm{form}}} \right)$

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,\left[ {{e^x} + {e^x} + x\,{e^x} + \frac{1}{{{{(1 + x)}^2}}}} \right]$

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,[1 + 1 + 0 + 1] = \frac{3}{2}$.

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