JEE PYQ: Limits Question 21
Question 21 - 2019 (08 Apr Shift 1)
$\lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$ equals:
(1) $4\sqrt{2}$
(2) $\sqrt{2}$
(3) $2\sqrt{2}$
(4) $4$
Show Answer
Answer: (1)
Solution
$= \lim_{x\to 0}\frac{\sin^2 x}{\sqrt{2} - \sqrt{2\cos^2\frac{x}{2}}} = \lim_{x\to 0}\frac{\sin^2 x}{\sqrt{2}(1 - \cos\frac{x}{2})} = \lim_{x\to 0}\frac{\sin^2 x}{\sqrt{2} \cdot 2\sin^2\frac{x}{4}} \cdot \frac{(\frac{\sin x}{x})^2 \cdot 16}{2\sqrt{2}(\frac{\sin(x/4)}{x/4})^2} = \frac{16}{2\sqrt{2}} = 4\sqrt{2}$.
BETA
