Maybe the logarithms can help

$\begin{aligned} L & = \lim_{x \to 0^+} \frac {e^{2x}}{x^x} \\ & = \lim_{x \to 0^+} \exp \left( \ln \left(\frac {e^{2x}}{x^x} \right) \right) \\ & = \exp \left(\lim_{x \to 0^+} \ln \left(\frac {e^{2x}}{x^x} \right) \right) \\ & = \exp \left(\lim_{x \to 0^+} \left(2x - x \ln x \right) \right) \\ & = \exp \left(\lim_{x \to 0^+} \left(- x \ln x \right) \right) \\ & = \exp \left(\lim_{x \to 0^+} \left(-{\color{#3D99F6}\frac{\ln x}{\frac 1x}} \right) \right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{x \to 0^+} \left({\color{#3D99F6}\frac{\frac 1x}{\frac 1{x^2}}} \right) \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \exp \left(\lim_{x \to 0^+} x \right) \\ & = e^0 = \boxed{1} \end{aligned}$