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 .
$$=\lim_{x\to 0}\frac{1-\sin^2 x -\sqrt{\cos x}}{x^2}=\lim_{x\to 0}\frac{1-\cos x}{x^2(1+\sqrt{\cos x})}-\lim_{x\to 0}\frac{\sin^2 x}{x^2}=\frac{1}{2}\lim_{x\to 0}\frac{1-\cos x}{x^2}-1.$$ Can you continue?
Method $\#1:$
$\displaystyle\lim_{x\to0}\dfrac{1-\cos^nx}{x^2}=\lim_{x\to0}\dfrac{\sum_{r=0}^{n-1}\cos^rx}{1+\cos x}\cdot\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2=\dfrac n2$
$\displaystyle\lim_{x\to0}\dfrac{1-\cos^{1/m}x}{x^2}=\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2\cdot\lim_{x\to0}\dfrac1{(1+\cos x)\sum_{r=0}^{1/m-1}\cos^rx}=\dfrac1{2m}$
Now $\dfrac{\cos^2x-\sqrt{\cos x}}{x^2}=\dfrac{1-\cos^{1/2}x}{x^2}-\dfrac{1-\cos^2x}{x^2}$
So, here $n=m=2$
Method $\#2:$ Using Binomial series, for finite $n,$
$$1-\cos^n(2y)=1-(1-2\sin^2y)^n=-\sum_{r=1}^n\binom nr(-2\sin^2y)^r$$
$$\implies\lim_{y\to0}\dfrac{1-\cos^n(2y)}{y^2}=\left(\lim_{y\to0}\dfrac{\sin y}y\right)^2\cdot\lim_{y\to0}\left[-\binom n1(-2)-\sum_{r=1}^n\binom nr(-2)^r\sin^{2r-2}\right]=2n$$
$$\frac{\cos^2x-\sqrt{\cos x}}{x^2}=\frac{\cos^4x-\cos x}{x^2(\cos^2x+\sqrt{\cos x})}=\frac{\cos x}{\cos^2x+\sqrt{\cos x}}\frac{\cos^3x-1}{x^2}=\frac{\cos x}{\cos^2x+\sqrt{\cos x}}\frac{(\cos x-1)(\cos^2 x+\cos x+1)}{x^2}=\frac{\cos x(\cos^2 x+\cos x+1)}{\cos^2x+\sqrt{\cos x}}\frac{(\cos x-1)}{x^2}$$
Now, use $\cos 2x=1-2\sin^2 x \implies 2\sin^2 x=1-\cos 2x$ or $2\sin^2 (x/2)=1-\cos(x).$
Two standard limits $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1},\,\lim_{x\to 0}\frac{1-\cos x} {x^{2}}=\frac{1}{2}$$ come to our rescue here. We have \begin{align} L&=\lim_{x\to 0}\frac{\cos^{2}x-\sqrt{\cos x}} {x^{2}}\notag\\ &=\lim_{x\to 0}\frac{\cos x - 1}{x^{2}}\cdot\left(\frac{\cos^{2}x-1}{\cos x - 1}-\frac{\cos^{1/2}x - 1}{\cos x - 1}\right)\notag\\ &=-\frac{1}{2}\left(2-\frac{1}{2}\right)\notag\\ &=-\frac{3}{4}\notag \end{align}
