Finding $\displaylines{\lim_{x\to 0}\frac {1-\sqrt{\cos x}}{x^2}}$. [duplicate]

$\displaylines{\lim_{x\to 0}\frac {1-\sqrt {\cos x}}{x^2}}$.

This is how I've approached this problem:

$\displaylines{\lim_{x\to 0}\frac {1-\sqrt{\cos x}}{x^2}=\lim_{x\to 0}\frac{(1-\sqrt{\cos x})(1+\sqrt{\cos x})}{x^2(1+\sqrt{\cos x})}=\lim_{x\to 0}\frac{1-\cos x}{x^2(1+\sqrt{\cos x})}=\lim_{x\to 0}\frac{\sin^2\frac{x}{2}}{x^2(1+\sqrt{\cos x})}=\frac{1}{2}\lim_{x\to 0}\frac{(\sin\frac{x}{2})^2}{x^2}\\=\frac{1}{2}\lim_{x\to 0}\left (\frac{\sin\frac{x}{2}}{2\frac{x}{2}}\right )^2=\frac{1}{2}\lim_{x\to 0}\left (\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right )^2 \frac{1}{4} =\frac{1}{8}}$.

However, the result in my textbook is $\frac{1}{4}$. Am I missing something?

Solution 1:

I'm very embarrased, you and your book are both wrong... $\frac14$ is the answer.

$$ \lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \lim_{x\to0}\frac{(1-\sqrt{\cos x})(1+\sqrt{\cos x})}{x^2(1+\sqrt{\cos x})} = \lim_{x\to0}\frac{1-\cos x}{x^2}\lim_{x\to0}\frac1{1+\sqrt{\cos x}} \\ = \frac12\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)} = \frac12\lim_{x\to0}\frac{\sin^2x}{x^2}\lim_{x\to0}\frac1{1+\cos x} \\ = \frac14 $$