Evaluate the limit $\lim_{x\to 0} \frac{x-\sin x - \cos^{-1} (e^{-\frac x2})}{x^2}$

Question

The limit becomes $$\lim_{x\to 0} \frac{x-\sin x -0}{x^2}$$ $$=\lim_{x\to 0} \frac{1-\cos x}{2x}$$ $$=0$$

$0$ is not the right answer. I think my mistake was in the first step, but I couldn’t think of anything else.

you can't just evaluate the arccos to 0, you still need to take derivatives of it — user29418, Oct 07 '20 at 17:45

user · Accepted Answer · 2020-10-07T20:30:38.823

4

We have that as $x\to 0^+$

$$\cos^{-1} (e^{-\frac x2})=\sqrt x -\frac1{12}x^{\frac32}+o(x^2)$$

therefore the right limit is $-\infty$.

The left limit doesn’t exist since $e^{-x/2}\to 1^+$.

Refer to the related

As an alternative since

$$\frac{x-\sin x}{x^2}\to 0$$

by $-\frac x2=\log(\cos y)$ as $y\to 0^+$ we have

$$\frac{-\cos^{-1}\left(e^{-\frac x2}\right)}{x^2}=-\frac{y}{4\log^2\cos y}$$

which can be easily solved by l’Hospital.

edited Oct 07 '20 at 20:30

answered Oct 07 '20 at 17:29

user

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How did you write the expansion? did you use the expansion of arccos and then $e^{-\frac{x}{2}}$ – Anindya Prithvi Oct 07 '20 at 17:33
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I add the source for that! – user Oct 07 '20 at 17:37
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@user Another way (rather strenous) is to use $t=e^x$ substituition and then use L-Hospitals rule twice to get $- \infty$.But your method is better! – Albus Dumbledore Oct 07 '20 at 17:49
Is there any way other an expansion, since the expansion can’t always be easily determined? L Hospital is way too tedious – Aditya Oct 07 '20 at 18:02
@Aditya Yes we can simplify that by substitution. I add something on that. – user Oct 07 '20 at 19:18

1 Answers1