Question

\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\] 

विकल्प

•  2/3 

• 4/3 

• \[- 2\sqrt{3}\] 

• −4/3

Answer 1

4/3 

\[\lim_{h \to 0} 2\left[ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h\left( \sqrt{3}\cos h - \sin h \right)} \right]\]
\[ = \lim_{h \to 0} 2\frac{\left[ \frac{\sqrt{3}}{2}\cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2}\cos h + \frac{\sin h}{2} \right]}{h\left( 3 \cos h - \sqrt{3} \sin h \right)}\]
\[ = \lim_{h \to 0} 2\left( \frac{2 \sin h}{h} \right) \times \frac{1}{\left( 3 \cos h - \sqrt{3}\sin h \right)}\]
\[ = \lim_{h \to 0} \frac{4}{3 \cos h - \sqrt{3} \sin h}\]
\[ = \frac{4}{3}\]