# 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.

• Why did you write $\sin\left(\frac{\sqrt{x^{2}}}{2}\right)$ instead of $\sin \frac{x}{2}$? $x$ is necessarily nonnegative here. Oct 25, 2019 at 9:35
• that's right,my mistake. Oct 25, 2019 at 9:46
• Fundamentally, it doesn't cha,ge the computation, but it makes it lighter. Oct 25, 2019 at 9:50

Using the standard Taylor expansion of $$\cos(x)$$, you should have $${\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}=\frac{\frac{x^2}{4}+O\left(x^4\right) } {\frac{x}{2}+O\left(x^2\right) }=\frac{x}{2}+O\left(x^2\right)$$

• The big gun always works fine!
– user
Oct 25, 2019 at 9:25

We have that

$$\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}= \frac{1-\sqrt{\cos\left(x\right)}}{x}\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}} \frac{x}{1-\cos\left(\sqrt{x}\right)}=$$

$$=x\frac{1-\cos\left(x\right)}{x^2}\frac{1}{1+\sqrt{\cos\left(x\right)}} \frac{x}{1-\cos\left(\sqrt{x}\right)}\to 0$$

$$\frac{1- \cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}=2x\cdot\frac{1- \cos\left(x\right)}{x^2}\frac{\left(\frac{\sqrt{x}}{2}\right)^2}{\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\to 0$$

Using $$\operatorname{sinc}y:=\frac{\sin y}{y}$$ so $$\lim_{y\to0}\operatorname{sinc}y=1$$, the last expression you obtained is$$\frac12\lim_{x\to0}\frac{x\operatorname{sinc}^2\frac{\sqrt{x^2}}{2}}{\operatorname{sinc}^2\frac{\sqrt{x}}{2}}=\frac12\lim_{x\to0}\frac{x\cdot1}{1}=0.$$

• I had seen $\textrm{sinc}$ written in some notes but I absolutely did not remember that it referred to $\sin y/y$. 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 $$\dfrac{x^2}2$$.

• Maybe it should be pointed out that It is true in this specific case but it is not true in general.
– user
Oct 25, 2019 at 9:25
• @user: it is true in general.
– user65203
Oct 25, 2019 at 9:25
• For example $\frac{\sin x-x}{x^3}\to 0$?
– user
Oct 25, 2019 at 9:27
• @user: this is not what I call "in a ratio", obviously. The sine appears in a subtraction.
– user65203
Oct 25, 2019 at 9:34
• 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.
– user
Oct 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 $$\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}=\dfrac{\lim_{x\to0}\dfrac{\sin x}x}{\dfrac{\lim_{x\to0}\sin(\sqrt x)}{\sqrt x}}\cdot\lim_{x\to0}\dfrac x{\sqrt x}=?$$