TITLE: Calculate $\lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}}$ without using L'Hôpital's rule.
QUESTION [5 upvotes]: Question:
Calculate 
$$\lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}}$$
without using L'Hôpital's rule.
Attempted solution:
A spontaneous substitution of t = $\arccos{x}$ gives:
$$\lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}} = \lim_{t \to 0^{+}} \frac{t}{\sqrt{1-\cos{t}}}$$
Using the half-angle formula for $\sin \frac{t}{2}$:
$$\lim_{t \to 0^{+}} \frac{t}{\sqrt{1-\cos{t}}} = \lim_{t \to 0^{+}} \frac{t}{\sqrt{2 \sin^{2}{(\frac{t}{2})}}} = \lim_{t \to 0^{+}} \frac{t}{\sqrt{2}\sin{(\frac{t}{2})}}$$
Forcing a standard limit:
$$\lim_{t \to 0^{+}} \frac{t}{\sqrt{2}\sin{(\frac{t}{2})}} = \lim_{t \to 0^{+}} \frac{\frac{t}{2}}{\frac{\sqrt{2}}{2}\sin{(\frac{t}{2})}} = \frac{2}{\sqrt{2}}$$
However, this is not correct as the limit is $\sqrt{2}$. Where have I gone wrong?

REPLY [2 votes]: Here is another approach that relies only on the Squeeze Theorem and the elementary inequalities from geometry
$$x\cos x\le \sin x\le x \tag 1$$
for $0\le x\le \pi/2$.  
Letting $x=\arccos(y)$ in $(1)$ reveals 
$$y\arccos(y)\le \sqrt{1-y^2}\le \arccos(y) \tag 2$$
for $y\le 1$.  
After rearranging $(2)$, we obtain for $0

$$\sqrt{1-y^2}\le \arccos(y)\le \frac{\sqrt{1-y^2}}{y} \tag 3$$
Now, dividing $(3)$ by $\sqrt{1-y}$, we have for $y<1$
$$\sqrt{1+y}\le \frac{\arccos(y)}{\sqrt{1-y}}\le \frac{\sqrt{1+y}}{y} \tag 4$$
Finally, applying the squeeze theorem to $(4)$ gives the expected result
$$\lim_{y\to 1^{-}}\frac{\arccos(y)}{\sqrt{1-y}}=\sqrt 2$$
And we are done!