Question

$\underset{h\to 0}{\lim }2\frac{\sqrt{3}\sin \left({\displaystyle \frac{\mathrm{\pi }}{6}}+h\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{6}}+h\right)}{\sqrt{3}h(\sqrt{3}\cos \left(h\right)-\sin \left(h\right))}=$

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

Answer 1

The correct option is C

$\frac{4}{3}$

Simplify the given function using trigonometric identities and limit:

$$\begin{gathered} \underset{h\to 0}{\lim }2\frac{\sqrt{3}\sin \left({\displaystyle \frac{\mathrm{\pi }}{6}+h}\right)-\cos \left({\displaystyle \frac{\mathrm{\pi }}{6}+h}\right)}{\sqrt{3}h(\sqrt{3}\cos \left(h\right)-\sin \left(h\right))} \\ =\underset{h\to 0}{\lim }2\frac{2\left[\frac{\sqrt{3}}{2}\sin \left(\frac{\mathrm{\pi }}{6}+h\right)-\frac{1}{2}\cos \left(\frac{\mathrm{\pi }}{6}+h\right)\right]}{2\sqrt{3}h\left(\frac{\sqrt{3}}{2}\cos \left(h\right)-\frac{1}{2}\sin \left(h\right)\right)} \\ =\underset{h\to 0}{\lim }2\frac{{\displaystyle \left[\sin \left(\frac{\mathrm{\pi }}{6}+h\right)\cos \left(\frac{\mathrm{\pi }}{6}\right)-\cos \left(\frac{\mathrm{\pi }}{6}+h\right)\sin \left(\frac{\mathrm{\pi }}{6}\right)\right]}}{{\displaystyle \sqrt{3}h\left(\cos \left(\frac{\mathrm{\pi }}{6}\right)\cos \left(h\right)-\sin \left(\frac{\mathrm{\pi }}{6}\right)\sin \left(h\right)\right)}} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle 2\sin \left(h\right)}}{{\displaystyle \sqrt{3}h\left(\cos \left(\frac{\mathrm{\pi }}{6}+h\right)\right)}} \\ =\frac{2}{\sqrt{3}}\times \frac{2}{\sqrt{3}} \\ =\frac{4}{3} \end{gathered}$$

Therefore, Option(C) is the correct answer.