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$$ \lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}} $$

I was doing some questions on limits, I saw one in which there is $\arccos x$. I am stuck there, not able to proceed.

Can you give me some hint?

Travis Willse
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akbar
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  • when $x$ tends to $\frac1{\sqrt 2}$ from left we get $-\infty$ and when from right we get $+\infty$ and hence limit doesn't exist.

    i mean LHL$\ne$RHL

     – user5954246 May 26 '16 at 14:58
  • @DeNiSkA how can we say about LHL AND RHL – akbar May 26 '16 at 16:13
  • when i say $x$ tends to $\frac1{\sqrt2}$ i mean LHL and when i say from right i mean RHL, i hope this makes some sense! – user5954246 May 26 '16 at 17:21
  • @DeNiSkA but how can we say it is -∞ or +∞. – akbar May 26 '16 at 17:25
  • limit of $1/x$ doesn't exist at $x=0$ why because LHL$\ne$RHL, similarly as we know that numerator of your limit tends to a finite number and denominator tends to two different numbers so it doesn't exist – user5954246 May 26 '16 at 18:49
  • you many give me a counter example of $\sin x/x$ at $x=0$ but when denomiator tends to a negative number even numerator tends to negative number hence limit exists, is it making sense? – user5954246 May 26 '16 at 18:50

3 Answers3

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Set $t=\arccos(2x\sqrt{1-x^2})$, so $2x\sqrt{1-x^2}=\cos t$ and, by definition, $0\le t\le\pi$. Then $$ \sin t=\sqrt{1-\cos^2t}=\sqrt{1-4x^2+4x^4}=|2x^2-1| $$ and so $t=\arcsin|2x^2-1|$. Thus your limit (from the right) can be written $$ \lim_{x\to(1/\sqrt2)^+}\sqrt{2}\frac{\arcsin|2x^2-1|}{\sqrt{2}x-1}= \lim_{x\to(1/\sqrt2)^+}\sqrt{2} \frac{(\sqrt{2}x+1)\arcsin(2x^2-1)}{2x^2-1} $$ which you should be able to compute.

Note then that, for $x<1/\sqrt{2}$, $|2x^2-1|=-(2x^2-1)$ and you can go similarly for the limit as $x\to(1/\sqrt{2})^-$

egreg
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First of all, $$\arccos2x\sqrt{1-x^2}=\dfrac\pi2-\arcsin2x\sqrt{1-x^2}$$

Now using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $, $$2\arcsin x=\begin{cases}\arcsin2x\sqrt{1-x^2} &\mbox{if } |x|\le\dfrac1{\sqrt2} \\ \pi-\arcsin2x\sqrt{1-x^2} & \mbox{if } x>\dfrac1{\sqrt2}\\-\pi-\arcsin2x\sqrt{1-x^2} & \mbox{if } x<-\dfrac1{\sqrt2} \end{cases}$$

$$\implies\lim_{x\to\frac1{\sqrt2}^+}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=2\lim_{x\to\frac1{\sqrt2}^+}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=\dfrac{d(\arcsin x)}{dx}_{(\text{ at } x=\pi/4)}$$

and $$\lim_{x\to\frac1{\sqrt2}^-}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=-2\lim_{x\to\frac1{\sqrt2}^-}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=?$$

Community
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lab bhattacharjee
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That $\sqrt{1-x^2}$ shouts "Trigonometric functions!!!". So:

Put $x=\sin\theta$, which means that $\sqrt{1-x^2}=\cos\theta$. Then $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\arccos \left(\sin 2\theta \right)}{\sin \theta-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\frac{\pi}{2}-2\theta}{\sin \theta-\frac{1}{\sqrt{2}}}$$

You can use l'Hôpital from here. Or...

Put $\phi=\theta-\frac{\pi}{4}$. The limit becomes $$\lim_{\phi \to 0} \frac{-2\phi}{\sin \phi \cos\frac{\pi}{4} + \cos \phi \sin\frac{\pi}{4}-\frac{1}{\sqrt{2}}} \equiv \lim_{\phi \to 0} \frac{-2\phi}{\frac{1}{\sqrt{2}}(\sin \phi + \cos \phi -1)}$$

…which you can l'Hôpital again, or note that for small $\phi$, $\sin\phi$ is practically $\phi$ and $\cos\phi$ is practically 1, so that the limit becomes $$\frac{-2}{\frac{1}{\sqrt{2}}}$$ …which I leave to you to work out. This is essentially reproducing what l'Hôpital does, but by doing it yourself you can actually see what you are doing.

Martin Kochanski
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  • I think the limit does not exist – akbar May 26 '16 at 14:31
  • Over the range in question it makes no difference. – Martin Kochanski May 26 '16 at 14:32
  • when you use l'hopital rule it says limit doesn't exist so you can say nothing about original limit! – user5954246 May 26 '16 at 14:33
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    @akbar: if you are in doubt, get a calculator and work out the values of the expression for $(1/\sqrt 2)+0.1$, $(1/\sqrt 2)+0.01$, $(1/\sqrt 2)+0.001$. This is actually a very good way to "get the feel" of a question before solving it. – Martin Kochanski May 26 '16 at 14:34
  • The usual definition of $\arccos$ only takes nonnegative values, so we cannot have $\arccos(\ast) = \frac{\pi}{2} - 2 \theta$ for $\theta > \frac{\pi}{4}$, as the r.h.s. is negative for such values. – Travis Willse May 26 '16 at 14:38