Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^{2}}$

 

Solution:

$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^{2}}$

$=\lim _{y \rightarrow 0} \frac{2-\sqrt{3} \cos \left(y+\frac{\pi}{6}\right)-\sin \left(y+\frac{\pi}{6}\right)}{y^{2} \times 36}$

$=\frac{1}{36} \times \lim _{y \rightarrow 0} \frac{2-\frac{3}{2} \cos y+\frac{\sqrt{3}}{2} \sin y-\frac{1}{2} \cos y-\frac{\sqrt{3}}{2} \sin y}{y^{2}}$

$=\frac{1}{36} \times \lim _{y \rightarrow 0} \frac{2(1-\cos y)}{y^{2}}$

$=2 \times \frac{1}{2} \times \frac{1}{36}$

$=\frac{1}{36}$

$\therefore \lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^{2}}=\frac{1}{36}$

 

Leave a comment

None
Free Study Material