Hi everyone, we are asked to evaluate $\displaystyle \lim _{x \to 0^+}\(ln(x+1))^{1/x}$. Now what I did is I set $\displaystyle y=(ln(x+1))^{1/x}$ then took the natural log of both sides to bring down the exponent 1/x: $\displaystyle

\lim _{x \to 0^+}y=\lim _{x \to 0^+}\frac{ln(ln(x+1))}_x$. From there on I can't seem to figure out how to make the function in the form of 0/0 or infinity/infinity to use L'Hopital's rule. Any one got any ideas? Thanks y'all in advance!