First of all, you don't need l'Hopital's rule.

Secondly, to find the limit of $f(x)^g(x)$ (where $\lim \limits_{x \to 0} f(0)=\lim \limits_{x \to 0} g(0)=0$), you will write it as $\mathrm e^{g(x) \log f(x)}$. Thus we seek functions $f$ and $g$ such that $\lim \limits_{x \to 0} g(x) \log f(x) = k$. Selecting $f(x) = x$ for simplicity, we see that $g(x) = {k \over \log x}$ will do the job, because the $\log x$ terms will cancel for all non-zero $x$.

**Solution:**
$x^{k \over \log x}$