_x0 cosec,x(√2^2x+3 x-^2x+ x+4) is:

Question

$\displaystyle\lim_{x	o0}\ 	ext{cosec}\,x\left(\sqrt{2\cos^2x+3\cos x}-\sqrt{\cos^2x+\sin x+4}ight)$ is:

Accepted Answer

Rationalize by multiplying and dividing by the conjugate sum. The difference of the radicands is
$$\left(2\cos^2x+3\cos x\right)-\left(\cos^2x+\sin x+4\right)=\cos^2x+3\cos x-\sin x-4=(\cos x+4)(\cos x-1)-\sin x$$

So the expression becomes
$$\lim_{x\to0}\frac{(\cos x+4)(\cos x-1)-\sin x}{\sin x\left(\sqrt{2\cos^2x+3\cos x}+\sqrt{\cos^2x+\sin x+4}\right)}$$

Writing $\cos x-1=-2\sin^2\dfrac{x}{2}$ and $\sin x=2\sin\dfrac x2\cos\dfrac x2$, and cancelling a factor of $2\sin\dfrac x2$ from numerator and denominator:

$$\lim_{x\to0}\frac{-\left(\sin\frac x2(\cos x+4)+\cos\frac x2\right)}{\cos\frac x2\left(\sqrt{2\cos^2x+3\cos x}+\sqrt{\cos^2x+\sin x+4}\right)}$$

As $x\to0$, this tends to $\dfrac{-(0+1)}{1(\sqrt5+\sqrt5)}=\dfrac{-1}{2\sqrt5}$.

Solutions · 1