JEE Mains · Maths · STD 11 - 12. limits
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \,\cos \,x}}{{{{\sin }^2}\,x}}\) is equal to
- A \(2\)
- B \(3\)
- C \(\frac {3}{2}\)
- D \(\frac {5}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac {3}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathop {\lim }\limits_{x \to 0} \frac{{2x{e^{{x^2}}} + \sin x}}{{2\sin x\cos x}}\) \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sin x}}{e^{{x^2}}} + \frac{1}{2}} \right)\frac{1}{{\cos x}} = 1 + \frac{1}{2} = \frac{3}{2}\)
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