The value of $\lim _{h \rightarrow 0} 2\left\{\frac{\sqrt{3} \sin \left(\frac{\pi}{6}+h\right)-\cos \left(\frac{\pi}{6}+h\right)}{\sqrt{3} h(\sqrt{3} \cos h-\sin h)}\right\}$ is :

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Limits

The value of

lim_{h \to 0} 2 {\frac{3 s i n ( \frac{π}{6} + h ) - c o s ( \frac{π}{6} + h )}{3 h ( 3 c o s h - s i n h )}}

is :

Answer: (D)

The numerator collapses to a single sine:

3 sin (\frac{π}{6} + h) - cos (\frac{π}{6} + h) = 2 [sin (\frac{π}{6} + h) cos \frac{π}{6} - cos (\frac{π}{6} + h) sin \frac{π}{6}] = 2 sin h

so the whole expression is

2 \cdot \frac{2 sin h}{3 h ( 3 cos h - sin h )} = \frac{4}{3} \cdot \frac{sin h}{h} \cdot \frac{1}{3 cos h - sin h}

h \to 0

\frac{sin h}{h} \to 1

and

3 cos h - sin h \to 3

, so the limit is

\frac{4}{3} \cdot \frac{1}{3} = \frac{4}{3}

The value of the limit is

\frac{4}{3}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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