JEE PYQ: Limits Question 8
Question 8 - 2021 (26 Feb Shift 1)
The value of $\lim_{h \to 0} 2\left{\frac{\sqrt{3}\sin\left(\frac{\pi}{6} + h\right) - \cos\left(\frac{\pi}{6} + h\right)}{\sqrt{3}h(\sqrt{3}\cosh - \sinh)}\right}$ is:
(1) $\frac{3}{4}$
(2) $\frac{2}{3}$
(3) $\frac{4}{3}$
(4) $\frac{2}{3}$
Show Answer
Answer: (3)
Solution
$\lim_{h\to 0} 2 \times 2 \left{\frac{\sin\left(\frac{\pi}{6} + h - \frac{\pi}{6}\right)}{2\sqrt{3}h\cos\left(h + \frac{\pi}{6}\right)}\right} = \frac{2}{\sqrt{3}} \times \frac{2}{\sqrt{3}} = \frac{4}{3}$.
BETA
