Question

The value of
\(\displaystyle\lim_{h\to 0}\, 2\left\{\frac{\sqrt{3}\sin\left(\frac{\pi}{6}+h\right)-\cos\left(\frac{\pi}{6}+h\right)}{\sqrt{3}\,h\left(\sqrt{3}\cos h-\sin h\right)}\right\}\)
is :

• A. \(\frac{4}{3}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{3}{4}\)
• D. \(\frac{2}{\sqrt{3}}\)

Hint:

Write \(\sqrt{3}\sin x-\cos x = 2\sin\left(x-\frac{\pi}{6}\right)\) so the numerator becomes \(2\sin h\), then use \(\frac{\sin h}{h}\to 1\).

Answer 1

The Correct Option is A

Let \(x=\frac{\pi}{6}+h\). Then \(\sqrt{3}\sin x-\cos x = 2\left(\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x\right)=2\sin\left(x-\frac{\pi}{6}\right)=2\sin h\).
Numerator \(=2\sin h\).
As \(h\to 0\): \(\sqrt{3}\cos h-\sin h\to\sqrt{3}\), so the denominator \(\approx \sqrt{3}\,h\cdot\sqrt{3}=3h\).
Limit \(=2\cdot\lim_{h\to0}\frac{2\sin h}{3h}=2\cdot\frac{2}{3}=\frac{4}{3}\).
Answer: (1) \(\frac{4}{3}\)