The value of the limit lim x→0 (cosecx) (√((2cos^2x+3cosx)-√(cos^2x+sinx+4)) is

Question

The value of the limit

\(\lim _{x \rightarrow 0}(\operatorname{cosec} x)\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)\) is

(1) 0

(2) 1

(3) \(\frac{1}{2 \sqrt{5}}\)

(4) \(-\frac{1}{2 \sqrt{5}}\)
 

Answer 1

Correct option is (4) \(-\frac{1}{2 \sqrt{5}}\)   

After rationalization,

\(\lim _{x \rightarrow 0} \frac{1}{\sin x}\left(\frac{\left(2 \cos ^2 x+3 \cos x\right)-\left(\cos ^2 x+\sin x+4\right)}{\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}}\right)\)  

\(\lim _{x \rightarrow 0} \frac{\cos ^2 x+3 \cos x-\sin x-4}{(\sin x)(\sqrt{5}+\sqrt{5})}\)   

\(=\lim _{x \rightarrow 0} \frac{1}{2 \sqrt{5}} \frac{\cos ^2 x+3 \cos x-\sin x-4}{\sin x}\)  

L'Hopital,  

\(\Rightarrow \lim _{x \rightarrow 0}\left(\frac{1}{2 \sqrt{5}}\right)\left(\frac{2 \cos x(-\sin x)-3 \sin x-\cos x}{\cos x}\right) \)  

\(=\frac{1}{2 \sqrt{5}}\left[\frac{-1}{1}\right]=-\frac{1}{2 \sqrt{5}}\)