MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \,\cos \,x}}{{{{\sin }^2}\,x}}$ is equal to
  • A
    $2$
  • B
    $3$
  • $\frac {3}{2}$
  • D
    $\frac {5}{4}$

Answer

Correct option: C.
$\frac {3}{2}$
c
$\mathop {\lim }\limits_{x \to 0} \frac{{2x{e^{{x^2}}} + \sin x}}{{2\sin x\cos x}}$

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sin x}}{e^{{x^2}}} + \frac{1}{2}} \right)\frac{1}{{\cos x}} = 1 + \frac{1}{2} = \frac{3}{2}$

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