<s>[INST] Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate 
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}=  \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0}  \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$= 
  \lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
 [/INST] Only ${x\rightarrow 0^+}$ is possible here. Let $x=y^2$, then
$$L =\lim_{x \rightarrow 0^+} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}= \lim_{y \rightarrow 0} \frac{1-\sqrt{\cos y^2}}{1-\cos y} \lim_{y \rightarrow 0} \frac{1-\sqrt{1-y^4/2}}{1-(1-y^2/2)}= \lim_{y \rightarrow 0} \frac{1-(1-y^4/4)}{1-(1-y^2/2)}$$ $$\implies L= \lim_{y \rightarrow 0}\frac{y^4/4}{y^2/2}=\lim_{y \rightarrow 0} \frac{y^2}{2}=0.$$
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