Calcul of $\lim_{x\to 1^{-}}\frac{\pi-\arccos(x)}{\sqrt{1-x^2}}$ I found that
$$\lim_{x\to 1^{-}}\frac{\pi-\arccos(x)}{\sqrt{1-x^2}}=+\infty$$
My question: Can we use Taylor series method to find this limit?
 A: Note that we cannot use l’Hopital since as $x\to1^-$
$$\frac{\pi-\arccos(x)}{\sqrt{1-x^2}}$$
is in the form  $\frac{\pi}{0^+}$, and we can directly conclude from here that the limit is $+\infty$.
A more interesting limit would be the same for $x\to -1^+$ and in that case we can of course apply l'Hopital.
A: You cannot use Taylor series here, because $\arccos$ is not analytic at $1$. In fact, its not even continuous. You could try to do a kind of Taylor expansion with taking limits only from one side, but then the derivative at $1$ is still infinite.
There is a generalisation you can use though, the so called Puisseux Series.
As mentioned in the other answer, the more interesting limit is $x\to -1^+$.
As the first right derivative of $\arccos(x)$ is infinite, you cannot use the Taylor series here. However, l'hospital still works:
$$\lim_{x\to -1^{+}}\frac{\pi-\arccos(x)}{\sqrt{1-x^2}}=\lim_{x\to -1^{+}} \frac {-(1-x^2)^{-1/2}}{-x  (1-x^2)^{-1/2} } = \lim_{x\to -1^{+}}\frac 1 x =-1.$$
You could also try to use the mentioned puisseux series: However,
$$\pi -\arccos(x) =\sqrt2 \sqrt{x+1} + \frac {(x+1)^{\frac 3 2}}{6\sqrt 2}+ \dots$$
and $$ \sqrt{1-x^2}= \sqrt 2 \sqrt{x+1} - \frac {(x+1)^{\frac 3 2}}{2\sqrt 2}+\dots$$
and there does not seem to be a vanishing term in either of those. So it is actually less helpful than I thought at first.
