Calculate $\lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}}$ without using L'Hôpital's rule. 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?
 A: HINT: $$\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\sqrt{2}$$
A: Just to show a different way:
$$
\lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}} =
\lim_{t \to 0^{+}} \frac{t}{\sqrt{1-\cos{t}}}=
\lim_{t \to 0^{+}} \frac{t\sqrt{1+\cos{t}}}{\sqrt{1-\cos^2{t}}}=
\lim_{t\to0^+}\frac{t}{\sin t}\sqrt{1+\cos t}=\sqrt{2}
$$
A: 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!  
