MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{x + 2\,\sin \,x}}{{\sqrt {{x^2} + 2\sin \,x + 1} - \sqrt {{{\sin }^2}\,x - x + 1} }}$ is
- ✓$2$
- B$6$
- C$3$
- D$1$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{{x^2} + 2\sin x + 1 - {{\sin }^2}x - x + 1}}$ $\left( {\sqrt {{x^2} + 2\sin x + 1} + \sqrt {{{\sin }^2}x - x + 1} } \right)$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{x + 2\sin x}}{{{x^2} + 2\sin x - {{\sin }^2}x + x}}.\left( 2 \right)$
Applying $L'H$ Rule
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{2.\left( {1 + 2\cos x} \right)}}{{2x + 2\cos x - 2\sin x\cos x + 1}}$
$ = \frac{{2\left( 3 \right)}}{{2 + 1}} = 2$
Hence the correct answer is option $(A)$.
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