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Evaluate $$\lim_{x\to 1^{-}}(\arccos (x))^{1-x}$$ I've tried L'Hopetal rule, but derivative contains $$ln(arccos(x))$$ and it isn't defined in $x=1$. And because of that I couldn't use Taylor series

Dukalis
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3 Answers3

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Note that by letting $t=\arccos(x)$, $$\lim_{x\rightarrow 1^-}{(\arccos x)^{1-x}}= \lim_{t\rightarrow 0^+}{t^{1-\cos(t)}}= \lim_{t\rightarrow 0^+}\exp((1-\cos(t))\ln(t))\\= \lim_{t\rightarrow 0^+}\exp\left(\frac{t^2}{2}\ln(t)\right)=e^0=1.$$

Robert Z
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  • Why does ln(t) decrease slower than t^2? – Dukalis Dec 22 '16 at 22:20
  • Note that $\lim_{t\to 0^+}t\ln(t)=\lim_{s\to +\infty }1/s\ln(1/s)=\lim_{s\to +\infty }-\ln(s)/s=0$ because $s$ goes faster to infinity than $\ln(s)$. – Robert Z Dec 22 '16 at 22:22
  • Thank you very much! I got it – Dukalis Dec 22 '16 at 22:24
  • @Dukalis Well done! BTW the person who asked can mark one answer as "accepted" ;-). http://math.stackexchange.com/tour – Robert Z Dec 22 '16 at 22:25
  • Oh please, little children @Salah and RobertZ, if the op wants to accept an answer, they will. Stop begging the asker for points (accepts, votes or whatever). I'm writing this comment in part to draw attention to the pandering for rep you both display. – amWhy Dec 22 '16 at 23:04
  • In case you try to delete your comment which displays your pandering for rep, Robert Z, I'll quote your comment here: "BTW the person who asked can mark one answer as "accepted" ;-)" – amWhy Dec 22 '16 at 23:06
  • @amWhy I am sorry that you are disappointed. I am just quoting what is written here: http://math.stackexchange.com/tour Anyway, you are right, this action is deplorable. I will not ever leave a comment like that. I want to be a little GOOD child. – Robert Z Dec 23 '16 at 08:38
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Put $x=\cos(t)$.

when $x$ goes to $1^-, t $ goes to $0^+$.

thus

$$\lim_{t\to 0^+}(\arccos(\cos(t)))^{1-\cos(t)}$$

$$=\lim_{t\to 0^+}e^{(1-\cos(t))\ln(t) }$$

$$\lim_{t\to 0^+}e^{ \frac{1-\cos(t) }{t^2}t^2\ln(t)}=e^0=1.$$

hamam_Abdallah
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In THIS ANSWER, I showed that the arccosine function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\sqrt{1-t^2}\le \arccos(t)\le \frac{\sqrt{1-t^2}}{t} }\tag 1$$

for $0<t<1$.

Using $(1)$, we can write for $x\in (0,1)$

$$\left(1-x^2\,\right)^{(1-x)/2}\le (\arccos(x))^{1-x}\le \left(\frac{1-x^2}{x^2}\,\right)^{(1-x)/2} \tag 2$$

Recalling that $\lim_{x\to 1^-}(1-x)^{1-x}=\lim_{t\to 0^+}t^t=1$, and applying the squeeze theorem to $(2)$ yields that coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1^-}(\arccos(x))^{1-x}=1}$$

Community
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Mark Viola
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