Evaluate L = lim as x to 0 (2/pi arccos x)^(1/x)

Evaluate $\displaystyle L = \lim_{x\to0}\left(\frac{2}{\pi}\arccos x\right)^{1/x}$.

Step 1: Take the natural logarithm

Since $\arccos0 = \dfrac{\pi}{2}$, the base tends to $1$ while the exponent $\dfrac1x$ becomes infinite: the form is $1^{\infty}$. Write $L = e^{I}$ with

\[ I = \lim_{x\to0}\frac{\ln\left(\dfrac{2}{\pi}\arccos x\right)}{x}, \]

a quotient of the form $\dfrac00$, because $\ln1 = 0$.

Step 2: Apply l'Hôpital's rule

Differentiating the numerator by the chain rule, with $\bigl(\arccos x\bigr)' = -\dfrac{1}{\sqrt{1-x^{2}}}$, the constant $\dfrac{2}{\pi}$ cancels: \[\frac{d}{dx}\ln\left(\frac{2}{\pi}\arccos x\right) = \frac{1}{\dfrac{2}{\pi}\arccos x}\cdot\frac{2}{\pi}\cdot\left(-\frac{1}{\sqrt{1-x^{2}}}\right) = \frac{-1}{\arccos x\cdot\sqrt{1-x^{2}}},\]

while $\dfrac{d}{dx}\bigl[x\bigr] = 1$. Hence

\[ I = \lim_{x\to0}\frac{-1}{\arccos x\cdot\sqrt{1-x^{2}}}. \tag{1} \]

Step 3: Evaluate the exponent

The expression in (1) is continuous at $x = 0$, so substitute directly: \[I = \frac{-1}{\arccos0\cdot\sqrt{1-0}} = \frac{-1}{\dfrac{\pi}{2}\cdot1} = -\frac{2}{\pi}. \tag{2}\]

Step 4: Recover $L$

\[ L = e^{I} = e^{-2/\pi}. \]

Step 5: Verification

\[x = 0.01:\ 0.5279966; \qquad x = -0.01:\ 0.5301409; \qquad x = 0.001:\ 0.5289705 \;\;\checkmark\]

\[\text{Series: } \arccos x = \frac{\pi}{2}-x-\frac{x^{3}}{6}-\cdots \;\Longrightarrow\; \frac{2}{\pi}\arccos x = 1-\frac{2x}{\pi}+O\!\left(x^{3}\right) \;\;\checkmark\]

\[\ln\left(1-\frac{2x}{\pi}+\cdots\right) = -\frac{2x}{\pi}+O\!\left(x^{2}\right) \;\Longrightarrow\; \frac{\ln(\cdots)}{x}\to-\frac{2}{\pi} \;\;\checkmark\]