Finding value of limit involving cosine I know we can use Maclaurin expansion and l'hopital's rule for solving it but I want another way . Find value of $\lim_{x \to 0} \frac{\cos^2x - \sqrt{\cos x}}{x^2}$.
My try : I multiplied and then divided by conjugate of numerator but it didn't help .

$$\frac{\cos^2x-\sqrt{\cos x}}{x^2}=\frac{\cos^4x-\cos x}{x^2(\cos^2x+\sqrt{\cos x})}=\frac{\cos x}{\cos^2x+\sqrt{\cos x}}\frac{\cos^3x-1}{x^2}=\frac{\cos x}{\cos^2x+\sqrt{\cos x}}\frac{(\cos x-1)(\cos^2 x+\cos x+1)}{x^2}=\frac{\cos x(\cos^2 x+\cos x+1)}{\cos^2x+\sqrt{\cos x}}\frac{(\cos x-1)}{x^2}$$
Now, use $\cos 2x=1-2\sin^2 x \implies 2\sin^2 x=1-\cos 2x$ or $2\sin^2 (x/2)=1-\cos(x).$
