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?
Best Answer
HINT: $$\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\sqrt{2}$$