What is the $\lim_{x\to 1^{-}}(\arccos (x))^{1-x}$? 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
 A: Note that by letting $t=\arccos(x)$,
$$\lim_{x\rightarrow 1^-}{(\arccos x)^{1-x}}=
\lim_{t\rightarrow 0^+}{t^{1-\cos(t)}}=
\lim_{t\rightarrow 0^+}\exp((1-\cos(t))\ln(t))\\=
\lim_{t\rightarrow 0^+}\exp\left(\frac{t^2}{2}\ln(t)\right)=e^0=1.$$
A: Put $x=\cos(t)$.
when $x$ goes to $1^-, t $ goes to $0^+$.
thus
$$\lim_{t\to 0^+}(\arccos(\cos(t)))^{1-\cos(t)}$$
$$=\lim_{t\to 0^+}e^{(1-\cos(t))\ln(t)  }$$
$$\lim_{t\to 0^+}e^{ \frac{1-\cos(t) }{t^2}t^2\ln(t)}=e^0=1.$$
A: 
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

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

