Limits
Trigonometric Limits
JEE Advanced 2015 Paper 1
Grade 11
Question:
$\lim_{x \to \pi/4} \frac{\sqrt{2} - \cos x - \sin x}{(4x - \pi)^2}$
(1) $\frac{1}{8}$
(2) $\frac{\sqrt{2}}{32}$
(3) $\frac{1}{4\sqrt{2}}$
(4) $\frac{1}{16\sqrt{2}}$
Step-by-Step Solution
Key Concept: Let $x = \pi/4 + h$, $h \to 0$. Then $\cos x + \sin x = \sqrt{2} \cos h$. Numerator $= \sqrt{2}(1 - \cos h) \approx \sqrt{2} \cdot h^2/2$. Denominator $= (4h)^2 = 16h^2$. Limit $= \sqrt{2}/(2 \cdot 16) = \sqrt{2}/32$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)