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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.

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    $\begingroup$ Please, write your question correctly at least. $\endgroup$ Commented Nov 2, 2020 at 20:24
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    $\begingroup$ Here is a formatting guide. $\endgroup$
    – J.G.
    Commented Nov 2, 2020 at 20:24
  • $\begingroup$ but how? It's my first time sorry. $\endgroup$
    – Zakaria
    Commented Nov 2, 2020 at 20:25
  • $\begingroup$ Do you know Taylor expansion ? $\endgroup$ Commented Nov 2, 2020 at 20:40
  • $\begingroup$ Are you allowed to use the L'Hopital Rule? $\endgroup$
    – player3236
    Commented Nov 2, 2020 at 20:42

5 Answers 5

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

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

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

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    $\begingroup$ Thank you guys. I got it. $\endgroup$
    – Zakaria
    Commented Nov 2, 2020 at 21:12
  • $\begingroup$ No problem, Zakaria! (: $\endgroup$
    – PinkyWay
    Commented Nov 2, 2020 at 21:17
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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 $f'(x)=0$

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  • $\begingroup$ Starting from the Taylor expansion of $\cos x$, why $\sqrt{\cos x}=1-\frac{x^2}{4}+o(x^2)$? $\endgroup$
    – Sebastiano
    Commented Nov 2, 2020 at 22:15
  • $\begingroup$ @Sebastiano $\sqrt{1+t}=1+t/2+o(t)$ $\endgroup$
    – egreg
    Commented Nov 2, 2020 at 22:16
  • $\begingroup$ 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. $\endgroup$
    – Sebastiano
    Commented Nov 2, 2020 at 22:19
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    $\begingroup$ @Sebastiano Use $t=-x^2+o(x^2)$, so $\sqrt{1+t}=1+\frac{1}{2}(-x^2/2+o(x^2))+o(x^2)=1-x^2/4+o(x^2)$ $\endgroup$
    – egreg
    Commented Nov 2, 2020 at 22:21
  • $\begingroup$ Ah ok, now I have understood. Thank you very much...again. $\endgroup$
    – Sebastiano
    Commented Nov 2, 2020 at 22:22
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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?

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  • $\begingroup$ Yes it's easy now. Thank you so mush! $\endgroup$
    – Zakaria
    Commented Nov 2, 2020 at 21:09
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$$\frac{\sin(\pi\sqrt{\cos x})}x=\frac{\sin(\pi(1-\sqrt{\cos x}))}x.$$

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

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

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  • $\begingroup$ I got it. Thank you! $\endgroup$
    – Zakaria
    Commented Nov 2, 2020 at 21:11
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Since you already received good answers for the limit iteself.

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 $$\frac{\pi ^2 \left(3456+\pi ^2\right)}{165888}\approx 0.20620$$

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