## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0^+}x^{1/\ln x}=e.$$
To solve this limit we will use the fact that the natural exponential and the natural logarithmic functions are inverse to each other i.e. that $e^{\ln a}=a.$ This yields $$x^{1/\ln x}=e^{\ln x^{1/\ln x}}=e^{\frac{1}{\ln x}\ln x}=e^1=e.$$ Now, putting this into the limit we get $$\lim_{x\to0^+}x^{1/\ln x}=\lim_{x\to0^+}e=e.$$