Evaluating $\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$

I don't know how can this be solved: $$\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$$ I've tried to multiply by $$\pi\sqrt{\cos x}$$: $$\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})}{x\pi\sqrt{\cos x}}$$

but we have here a hard problem $$\lim_{x\to0}(1/x)$$.

You think about the limit on the right and on the left but the exercise supposed to be solved by the limit in $$0$$ and not on the right or the left of it.

• Please, write your question correctly at least. – hamam_Abdallah Nov 2 '20 at 20:24
• Here is a formatting guide. – J.G. Nov 2 '20 at 20:24
• but how? It's my first time sorry. – Zakaria Nov 2 '20 at 20:25
• Do you know Taylor expansion ? – TheSilverDoe Nov 2 '20 at 20:40
• Are you allowed to use the L'Hopital Rule? – player3236 Nov 2 '20 at 20:42

We know that: $$\sin(\alpha)=\sin(\pi-\alpha)$$ and

$$\lim_{x\to 0}\frac{\sin(x)}x=1\quad\&\quad\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$$

\begin{aligned}\lim_{x\to 0}\frac{\sin(\pi\sqrt{\cos(x)})}x&=\lim_{x\to 0}\frac{\sin(\pi(1-\sqrt{\cos(x)}))}{\pi(1-\sqrt{\cos(x)})}\frac{(1-\sqrt{\cos(x)})}{x^2}\frac{1+\sqrt{\cos(x)}}{1+\sqrt{\cos(x)}}\pi x\\&=\lim_{x\to 0}\frac{\sin(\pi(1-\cos(x)))}{\pi(1-\sqrt{\cos(x)})}\frac{1-\cos(x)}{x^2}\frac{\pi x}{1+\sqrt{\cos(x)}}=0\end{aligned}

• Thank you guys. I got it. – Zakaria Nov 2 '20 at 21:12
• No problem, Zakaria! (: – Invisible Nov 2 '20 at 21:17

Note that for $$x\to 0$$, $$\dfrac {\sin(\pi\sqrt{\cos x})}{\pi\sqrt{\cos x}} \to \dfrac {\sin \pi}{\pi}=0$$. The limit for $$\dfrac {\sin x}x$$ is $$1$$ for $$x \to 0$$ only.

Hence we transform the argument of the sine so it tends to $$0$$, by seeing

$$\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x = \lim_{x\to0} \frac{\sin(\pi - \pi\sqrt{\cos x})} x = \lim_{x\to0} \frac{\sin(\pi - \pi\sqrt{\cos x})} {\pi-\pi\sqrt{\cos x}}\cdot \frac {\pi-\pi\sqrt{\cos x}}x$$

Now the first part of the product has limit $$1$$. Do you know how to calculate the limit of the second part?

• Yes it's easy now. Thank you so mush! – Zakaria Nov 2 '20 at 21:09

Since $$\cos x=1-x^2/2+o(x^2)$$, we also have $$\sqrt{\cos x}=1-\frac{x^2}{4}+o(x^2)$$ and so your limit can be rewritten as $$\lim_{x\to0}\frac{\sin(\pi-\pi x^2/4+o(x^2))}{x}=\lim_{x\to0}\frac{\sin(\pi x^2/4+o(x^2))}{x}=\lim_{x\to0}\frac{\pi x^2/4+o(x^2)}{x}=0$$ With $$x^2$$ at the denominator you'd get $$\pi/4$$.

A different way to solve the problem is to notice that the given limit is the derivative of $$f(x)=\sin(\pi\sqrt{\cos x})$$ at $$0$$; since $$f'(x)=\cos(\pi\sqrt{\cos x})\frac{-\pi\sin x}{2\sqrt{\cos x}}$$ we get $$f'(x)=0$$

• Starting from the Taylor expansion of $\cos x$, why $\sqrt{\cos x}=1-\frac{x^2}{4}+o(x^2)$? – Sebastiano Nov 2 '20 at 22:15
• @Sebastiano $\sqrt{1+t}=1+t/2+o(t)$ – egreg Nov 2 '20 at 22:16
• If $\cos x=1-x^2/2+o(x^2) \implies \sqrt{\cos x}=1-\frac{x^2}{4}+o(x^2)$, yet I have not understood after your comment. Excuse me very much. – Sebastiano Nov 2 '20 at 22:19
• @Sebastiano Use $t=-x^2+o(x^2)$, so $\sqrt{1+t}=1+\frac{1}{2}(-x^2/2+o(x^2))+o(x^2)=1-x^2/4+o(x^2)$ – egreg Nov 2 '20 at 22:21
• Ah ok, now I have understood. Thank you very much...again. – Sebastiano Nov 2 '20 at 22:22

$$\frac{\sin(\pi\sqrt{\cos x})}x=\frac{\sin(\pi(1-\sqrt{\cos x}))}x.$$

As the argument of the sine tends to $$0$$, we can bypass the sine and evaluate

$$\pi\lim_{x\to 0}\frac{1-\sqrt{\cos x}}x.$$

Now

$$\frac{1-\sqrt{\cos x}}x=\frac{1-\cos x}{(1+\sqrt{\cos x})x}=\frac{2\sin^2\dfrac x2}{(1+\sqrt{\cos x})x}\to0.$$

• I got it. Thank you! – Zakaria Nov 2 '20 at 21:11

You can go beyond the limit if you compose Taylor series (that is to say working one piece at the time).

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}+O\left(x^6\right)$$ $$\pi \sqrt{\cos(x)}=\pi -\frac{\pi x^2}{4}-\frac{\pi x^4}{96}+O\left(x^6\right)$$ $$\sin \left(\pi \sqrt{\cos (x)}\right)= \sin\left(\frac{\pi x^2}{4}+\frac{\pi x^4}{96}+O\left(x^6\right) \right)=\frac{\pi x^2}{4}+\frac{\pi x^4}{96}+O\left(x^6\right)$$ $$\frac{\sin(\pi\sqrt{\cos (x)})} x=\frac{\pi x}{4}+\frac{\pi x^3}{96}+O\left(x^5\right)$$ which shows the limit and how it is approached.

Moreover, this gives a shorcut evaluation of a not so pleasant expression. Suppose $$x=\frac \pi{12}$$ (this is quite far away from $$0$$. The exact expression would be $$\frac{\sin\left( \pi\sqrt{\frac {\sqrt 6+\sqrt 2} 4}\right)}{\frac \pi {12}} \approx 0.20612$$ while the above truncated series gives $$\frac{\pi ^2 \left(3456+\pi ^2\right)}{165888}\approx 0.20620$$