I'd like to evaluate the limit $$\lim_{x\to 0} \frac{\cos^{-1}(1-x)}{\sqrt{x}}$$
Actually I know that the limit equals $\sqrt{2}$, but I think the way I'm trying to evaluate it has a flaw.
$$\lim_{x\to 0} \frac{\cos^{-1}(1-x)}{\sqrt{x}} = \lim_{x\to 0} \frac{\frac{\pi}{2}- \sin^{-1}(1-x)}{\sqrt{x}} = \lim_{x\to 1} \frac{\frac{\pi}{2}- \sin^{-1}(x)}{\sqrt{1-x}} = \lim_{x\to 1} \frac{\frac{\pi}{2}- \sin^{-1}(x)}{1-x} \sqrt{1-x}$$
And we know that $\frac{d\sin^{-1}(x)}{dx}=\frac{1}{\sqrt{1-x^{2}}}$. But at the point $x=1$ it's not defined, so, theoretically speaking we couldn't apply the products of the limits in this case. But, if we could, it'd turn out to yield the right answer:
$$\lim_{x\to 1} \frac{\frac{\pi}{2}- \sin^{-1}(x)}{1-x} \sqrt{1-x} = \lim_{x\to 1} \frac{\frac{\pi}{2}- \sin^{-1}(x)}{1-x} \lim_{x\to 1}\sqrt{1-x}=\frac{1}{\sqrt{1-x^2}}\Bigr|_{x=1}\sqrt{x-1}\Bigr|_{x=1}=\sqrt{x+1}\Bigr|_{x=1} = \sqrt{2}$$
So If I'm right, I can't do this because the first limit would yield $\infty$. I would appreciate if someone could confirm wheteher I`m right, and the correct approach for this problem.
Thanks in advance.
Best Answer
For $\arccos{(1-x)}$ to be real, we need $x>0$, so replace $x=y^2$, and then since $\arccos{1}=0$, the limit becomes $$ \lim_{y \to 0} \frac{\arccos{(1-y^2)-\arccos{1}}}{y}, $$ which is the derivative of $\arccos{(1-y^2)}$ at $y=0$, i.e. $$ \left. +\frac{2y}{\sqrt{1-(1-y^2)^2}} \right|_{y=0} = \left. \frac{2y}{\sqrt{2y^2-y^4}} \right|_{y=0} = \left. \frac{2}{\sqrt{2-y^2}} \right|_{y=0} = \sqrt{2}. $$