_ x 0 e^ x^2 - x x^2 = - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = $

✓
$\frac{3}{2}$
B
$ - \frac{1}{2}$
C
$1$
D
None of these

✓

Answer

Correct option: A.

$\frac{3}{2}$

a
(a) $\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{{x^2}}} + \cos x}}{{x^2}}$

Now expanding ${e^{{x^2}}}$ and $\cos x,$ we get

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{{3{x^2}}}{{2\,!}} + {x^4}\,\left( {\frac{1}{{2\,!}} - \frac{1}{{4\,!}}} \right) + .......}}{{{x^2}}} = \frac{3}{2}$

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \,\frac{{2x{e^{{x^2}}} + \sin x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \,{e^{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x}}{{2x}} $

$= 1 + \frac{1}{2} = \frac{3}{2}.$

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