Question

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

(a) 2/3
(b) 4/3
(c) $-2\sqrt{3}$
(d) −4/3

Answer 1

(d) 4/3

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