Calculate $\lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}}$ without using L'Hôpital's rule.

Question

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?

Answer 1

HINT: $$\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\sqrt{2}$$

Answer 2

Just to show a different way: $$ \lim_{x \to 1^{-}} \frac{\arccos{x}}{\sqrt{1-x}} = \lim_{t \to 0^{+}} \frac{t}{\sqrt{1-\cos{t}}}= \lim_{t \to 0^{+}} \frac{t\sqrt{1+\cos{t}}}{\sqrt{1-\cos^2{t}}}= \lim_{t\to0^+}\frac{t}{\sin t}\sqrt{1+\cos t}=\sqrt{2} $$

Answer 3

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!