Limits
Trigonometric Rationalization
JEE Advanced 2020 Paper 1
Grade 11
Question:
$\lim_{x \to \pi/2} \frac{\sqrt{2 - \sqrt{1 + \sin x}}}{\cos^2 x}$
(1) $\frac{1}{4\sqrt{2}}$
(2) $\frac{\sqrt{2}}{8}$
(3) $\frac{1}{\sqrt{2}}$
(4) $\frac{\sqrt{2}}{4}$
Step-by-Step Solution
Key Concept: Let $x = \pi/2 - t$, $t \to 0$: $\sin x = \cos t$, $\cos^2 x = \sin^2 t$. Numerator $= \sqrt{2 - \sqrt{1 + \cos t}} \approx \sqrt{2 - \sqrt{2 - t^2/2}} \approx \sqrt{2 - \sqrt{2}(1 - t^2/4)}$. Rationalise to extract $t$ and simplify; result $= \sqrt{2}/8$.
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Correct Answer: (2)