Evaluate L = lim as x to infinity (pi/2- arctan x)^(1/ln

Evaluate $\displaystyle L = \lim_{x\to\infty}\left(\frac{\pi}{2}-\arctan x\right)^{1/\ln x}$.

Step 1: Take the natural logarithm

The base tends to $0$ and the exponent $\dfrac{1}{\ln x}$ tends to $0$ as well: the form is $0^{0}$. Write $L = e^{I}$ with

\[ I = \lim_{x\to\infty}\frac{\ln\left(\dfrac{\pi}{2}-\arctan x\right)}{\ln x}. \]

Step 2: Asymptotics of the base

For $x\gt 0$ the identity $\arctan x = \dfrac{\pi}{2}-\arctan\dfrac1x$ gives

\[ \frac{\pi}{2}-\arctan x = \arctan\frac1x, \]

and with $\arctan u = u-\dfrac{u^{3}}{3}+\cdots$ for small $u = \dfrac1x$: \[\frac{\pi}{2}-\arctan x = \frac1x-\frac{1}{3x^{3}}+O\!\left(x^{-5}\right). \tag{1}\]

Step 3: Take the logarithm of (1)

\[ \ln\left(\frac{\pi}{2}-\arctan x\right) = \ln\left(\frac1x\left[1-\frac{1}{3x^{2}}+O\!\left(x^{-4}\right)\right]\right) = -\ln x+\ln\left(1-\frac{1}{3x^{2}}+\cdots\right), \]

and since $\ln(1+u)\sim u$ for small $u$, \[\ln\left(\frac{\pi}{2}-\arctan x\right) = -\ln x-\frac{1}{3x^{2}}+O\!\left(x^{-4}\right). \tag{2}\]

Step 4: Compute the exponent

Dividing (2) by $\ln x$: \[I = \lim_{x\to\infty}\left[-1-\frac{1}{3x^{2}\ln x}+O\!\left(\frac{1}{x^{4}\ln x}\right)\right] = -1. \tag{3}\]

Step 5: Recover $L$

\[ L = e^{I} = e^{-1} = \frac1e. \]

Step 6: Verification

\[x = 100:\ 0.3678768; \qquad x = 10^{4}:\ 0.3678794; \qquad \tfrac1e = 0.3678794 \;\;\checkmark\]

\[\text{Dominant behaviour: } \left(\frac1x\right)^{1/\ln x} = e^{-\ln x/\ln x} = e^{-1} \;\;\checkmark\]

\[\text{Any base } \sim Cx^{-k} \text{ would give } e^{-k},\ \text{the constant } C \text{ being washed out by } \tfrac{1}{\ln x} \;\;\checkmark\]