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.
 A: Using the standard Taylor expansion of $\cos(x)$, you should have
$${\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}=\frac{\frac{x^2}{4}+O\left(x^4\right) } {\frac{x}{2}+O\left(x^2\right) }=\frac{x}{2}+O\left(x^2\right)$$
A: We have that
$$\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}=
\frac{1-\sqrt{\cos\left(x\right)}}{x}\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}
\frac{x}{1-\cos\left(\sqrt{x}\right)}=$$
$$=x\frac{1-\cos\left(x\right)}{x^2}\frac{1}{1+\sqrt{\cos\left(x\right)}}
\frac{x}{1-\cos\left(\sqrt{x}\right)}\to 0$$
By your way from here
$$\frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}=2x\cdot\frac{1-
\cos\left(x\right)}{x^2}\frac{\left(\frac{\sqrt{x}}{2}\right)^2}{\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\to 0$$
A: Using $\operatorname{sinc}y:=\frac{\sin y}{y}$ so $\lim_{y\to0}\operatorname{sinc}y=1$, the last expression you obtained is$$\frac12\lim_{x\to0}\frac{x\operatorname{sinc}^2\frac{\sqrt{x^2}}{2}}{\operatorname{sinc}^2\frac{\sqrt{x}}{2}}=\frac12\lim_{x\to0}\frac{x\cdot1}{1}=0.$$
A: Only ${x\rightarrow 0^+}$ is possible here. Let $x=y^2$, then
$$L =\lim_{x \rightarrow 0^+} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}= \lim_{y \rightarrow 0} \frac{1-\sqrt{\cos y^2}}{1-\cos y} \lim_{y \rightarrow 0} \frac{1-\sqrt{1-y^4/2}}{1-(1-y^2/2)}= \lim_{y \rightarrow 0} \frac{1-(1-y^4/4)}{1-(1-y^2/2)}$$ $$\implies L= \lim_{y \rightarrow 0}\frac{y^4/4}{y^2/2}=\lim_{y \rightarrow 0} \frac{y^2}{2}=0.$$
A: Easy with  equivalents: as $\sin u\sim_{u\to 0} u$, and equivalence is compatible with multiplication and division, we have
$$\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{\!\!2}\sim_{x\to 0}\left(\frac{\frac{\sqrt{x^{2}}}{2}}{\frac{\sqrt{x}}{2}}\right)^{\!\!2}=\bigl(\sqrt x\bigr)^2=x \qquad(\text{as } x\ge 0)$$
A: Hint:
In a limit to $0$, you can replace $\sin(x)$ by $x$ in a ratio because
$$\sin(x)=x\frac{\sin(x)}x.$$
And for this reason, you can replace $1-\cos(x)$ by $\dfrac{x^2}2$.
A: $$\lim_{x\to0}\dfrac{1-\sqrt{\cos x}}{1-\cos\sqrt x}$$
$$=\lim_{x\to0}\dfrac{1-\cos x}{1-\cos^2\sqrt x}\cdot\lim_{x\to0}\dfrac{1+\cos\sqrt x}{1+\sqrt{\cos x}}$$
$$=\left(\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}\right)^2\cdot\lim_{x\to0}\dfrac{1+\cos\sqrt x}{(1+\sqrt{\cos x})(1+\cos x)}$$
Now $\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}=\dfrac{\lim_{x\to0}\dfrac{\sin x}x}{\dfrac{\lim_{x\to0}\sin(\sqrt x)}{\sqrt x}}\cdot\lim_{x\to0}\dfrac x{\sqrt x}=?$
