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.
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$\begingroup$ Why did you write $\sin\left(\frac{\sqrt{x^{2}}}{2}\right)$ instead of $\sin \frac{x}{2} $? $x$ is necessarily nonnegative here. $\endgroup$– BernardOct 25, 2019 at 9:35
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$\begingroup$ that's right,my mistake. $\endgroup$– AbsurdOct 25, 2019 at 9:46
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$\begingroup$ Fundamentally, it doesn't cha,ge the computation, but it makes it lighter. $\endgroup$– BernardOct 25, 2019 at 9:50
7 Answers
Using the standard Taylor expansion of $\cos(x)$, you should have $${\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}=\frac{\frac{x^2}{4}+O\left(x^4\right) } {\frac{x}{2}+O\left(x^2\right) }=\frac{x}{2}+O\left(x^2\right)$$
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
$$\frac{1- \cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}=2x\cdot\frac{1- \cos\left(x\right)}{x^2}\frac{\left(\frac{\sqrt{x}}{2}\right)^2}{\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\to 0$$
Using $\operatorname{sinc}y:=\frac{\sin y}{y}$ so $\lim_{y\to0}\operatorname{sinc}y=1$, the last expression you obtained is$$\frac12\lim_{x\to0}\frac{x\operatorname{sinc}^2\frac{\sqrt{x^2}}{2}}{\operatorname{sinc}^2\frac{\sqrt{x}}{2}}=\frac12\lim_{x\to0}\frac{x\cdot1}{1}=0.$$
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$\begingroup$ I had seen $\textrm{sinc}$ written in some notes but I absolutely did not remember that it referred to $\sin y/y$. $\endgroup$ Oct 25, 2019 at 16:12
Only ${x\rightarrow 0^+}$ is possible here. Let $x=y^2$, then $$L =\lim_{x \rightarrow 0^+} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}= \lim_{y \rightarrow 0} \frac{1-\sqrt{\cos y^2}}{1-\cos y} \lim_{y \rightarrow 0} \frac{1-\sqrt{1-y^4/2}}{1-(1-y^2/2)}= \lim_{y \rightarrow 0} \frac{1-(1-y^4/4)}{1-(1-y^2/2)}$$ $$\implies L= \lim_{y \rightarrow 0}\frac{y^4/4}{y^2/2}=\lim_{y \rightarrow 0} \frac{y^2}{2}=0.$$
Easy with equivalents: as $\sin u\sim_{u\to 0} u$, and equivalence is compatible with multiplication and division, we have $$\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{\!\!2}\sim_{x\to 0}\left(\frac{\frac{\sqrt{x^{2}}}{2}}{\frac{\sqrt{x}}{2}}\right)^{\!\!2}=\bigl(\sqrt x\bigr)^2=x \qquad(\text{as } x\ge 0)$$
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 $\dfrac{x^2}2$.
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2$\begingroup$ Maybe it should be pointed out that It is true in this specific case but it is not true in general. $\endgroup$– userOct 25, 2019 at 9:25
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$\begingroup$ @user: it is true in general. $\endgroup$– user65203Oct 25, 2019 at 9:25
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$\begingroup$ @user: this is not what I call "in a ratio", obviously. The sine appears in a subtraction. $\endgroup$– user65203Oct 25, 2019 at 9:34
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$\begingroup$ For example $\frac{\frac{\sin x}x-1}{x^2}\to 0$? Sorry but I think that this hint, in that form, is quite confusing. Let say simply $\sin x=x+o(x)$ which avoids any possible confusion. $\endgroup$– userOct 25, 2019 at 9:38
$$\lim_{x\to0}\dfrac{1-\sqrt{\cos x}}{1-\cos\sqrt x}$$ $$=\lim_{x\to0}\dfrac{1-\cos x}{1-\cos^2\sqrt x}\cdot\lim_{x\to0}\dfrac{1+\cos\sqrt x}{1+\sqrt{\cos x}}$$
$$=\left(\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}\right)^2\cdot\lim_{x\to0}\dfrac{1+\cos\sqrt x}{(1+\sqrt{\cos x})(1+\cos x)}$$
Now $\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}=\dfrac{\lim_{x\to0}\dfrac{\sin x}x}{\dfrac{\lim_{x\to0}\sin(\sqrt x)}{\sqrt x}}\cdot\lim_{x\to0}\dfrac x{\sqrt x}=?$
