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$\frac {5}{4}$
$\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sin x}}{e^{{x^2}}} + \frac{1}{2}} \right)\frac{1}{{\cos x}} = 1 + \frac{1}{2} = \frac{3}{2}$
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