Evaluate L = lim as x to +0 (cot x)^(1/ln x)

Evaluate $\displaystyle L = \lim_{x\to+0}\bigl(\cot x\bigr)^{1/\ln x}$.

Step 1: Take the natural logarithm

As $x\to+0$: $\cot x\to+\infty$ and $\dfrac{1}{\ln x}\to0^{-}$, so the form is $\infty^{0}$. Write $L = e^{I}$ with

\[ I = \lim_{x\to+0}\frac{\ln\bigl(\cot x\bigr)}{\ln x}. \]

Step 2: Expand $\ln(\cot x)$

\[ \ln\bigl(\cot x\bigr) = \ln\bigl(\cos x\bigr)-\ln\bigl(\sin x\bigr). \tag{1} \]

From $\sin x = x\left(1-\dfrac{x^{2}}{6}+O\!\left(x^{4}\right)\right)$, \[\ln\bigl(\sin x\bigr) = \ln x+\ln\left(1-\frac{x^{2}}{6}+\cdots\right) = \ln x-\frac{x^{2}}{6}+O\!\left(x^{4}\right), \tag{2}\]

and from $\cos x = 1-\dfrac{x^{2}}{2}+O\!\left(x^{4}\right)$, \[\ln\bigl(\cos x\bigr) = -\frac{x^{2}}{2}+O\!\left(x^{4}\right). \tag{3}\]

Substituting (2) and (3) into (1): \[\ln\bigl(\cot x\bigr) = -\frac{x^{2}}{2}-\ln x+\frac{x^{2}}{6}+O\!\left(x^{4}\right) = -\ln x-\frac{x^{2}}{3}+O\!\left(x^{4}\right). \tag{4}\]

Step 3: Form the ratio

\[ I = \lim_{x\to+0}\frac{-\ln x-\dfrac{x^{2}}{3}+O\!\left(x^{4}\right)}{\ln x} = \lim_{x\to+0}\left(-1-\frac{x^{2}}{3\ln x}+\cdots\right) = -1, \tag{5} \]

since $\dfrac{x^{2}}{\ln x}\to0$.

Step 4: Recover $L$

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

Step 5: Verification

\[\cot x\sim\frac1x \;\Longrightarrow\; \bigl(\cot x\bigr)^{1/\ln x}\sim\left(\frac1x\right)^{1/\ln x} = e^{-\ln x/\ln x} = e^{-1} \;\;\checkmark\]

\[x = 0.1:\ 0.3684136; \qquad x = 0.01:\ 0.3678821; \qquad x = 0.001:\ 0.3678795 \;\;\checkmark\]

\[\text{The correction } -\frac{x^{2}}{3\ln x} \text{ is positive for } x\lt 1,\ \text{so the values approach from above} \;\;\checkmark\]