A not so easy limit

Using L'Hopital's (LH) rule and the Fundamental Theorem of Calculus (FToC), we have

$\begin{aligned} \lim_{n\to\infty} n^2 \int_0^{\frac{1}{n}} x^{x+1}\,dx &= \lim_{h\to 0^+} \frac{\int_0^h x^{x+1}\,dx}{h^2} & \left(\text{Setting } h = \frac{1}{n}\right) \\ &= \lim_{h\to 0^+} \frac{h^{h+1}}{2h} & \left(\text{LH and FToC}\right) \\ &= \frac{1}{2} \lim_{h\to 0^+} h^h \\ &= \frac{1}{2} \exp\left(\lim_{h\to 0^+} \frac{\ln h}{1/h}\right) & \left(\text{Continuity of } \exp(x) = e^x\right) \\ &= \frac{1}{2} \exp\left(\lim_{h\to 0^+} \frac{1/h}{-1/h^2}\right) & \left(\text{LH}\right) \\ &= \frac{1}{2} \exp\left(\lim_{h\to 0^+} (-h)\right) \\ &= \frac{1}{2} e^0 \\ &= \frac{1}{2} = \boxed{0.5} \end{aligned}$

Wow. I cant believe it was that simple... I turned the integral into a Riemann sum and solved a double limit that took me 30 minutes. Still got the right answer though.