Question

lim as x approaches 0 from the right and from the left of the x^(1/x)

Answer 1

\[\large \lim_{x \to 0^+} x^\frac{1}{x} \]
use the exponential trick, meaning rewrite it using the exponential function.
\[\large \lim_{x \to 0^+} e^{ \log ( x^\frac{1}{x})} \implies \lim_{ x \to 0^+} e^{\frac{1}{x} \log x} \]now use the fact that \(e\) is a continuous function, so you can switch the limit inside and rather analyze the following:
\[\large \lim_{x \to 0^+} \frac{1}{x} \log x \sim \infty \cdot -\infty \sim - \infty ' \]
back substitute the result into your expression to obtain \[\large \lim_{x \to 0^+} e ^{ \frac{1}{x}  \log x} = e^{- \infty'}=0\]

Answer 2

I have made use of asymptotic behavior of \( \log x\) , if you care for a different approach, use a taylor expression for \( \log x\) use that:
\[\large \log (1+x)= x- \frac{x^2}{2}+ \frac{x^3}{3}- \frac{x^4}{4}+ O(x^5) \] with a proper substitution, then solve analogous.

Answer 3

thank you so much!