Limit of a integral $\lim_{x\to 0+} \frac{1}{x^2} \int_{0}^{x} t^{1+t} dt$ Find the limit of 
$$\lim_{x\to 0+} \frac{1}{x^2} \int_{0}^{x} t^{1+t} dt$$. 
My idea to solve it is to use L'Hospital's rule but I am not sure why I can use it and how should i do it. Many thanks to them who are willing to help. 
 A: Formally, $t^{1+t}=te^{t\ln(t)}\to 0$ when $t\to 0^+$, because $\lim_{t\to 0^+} t\ln(t)=0$. It follows that, if $f(x)=\int_0^x t^{1+t}dt$, 
$$
\lim_{x\to 0^+}f(x)=0
$$
The function $f$ is differentiable at $0$. Having checked all the hypotheses for L'Hôspital's Rule, 
$$ 
\lim_{x\to 0^+} \frac{f(x)}{x^2}=\lim_{x\to 0^+} \frac{x^{1+x}}{2x}=\lim_{x\to 0^+} \frac{x^x}{2}=\frac{1}{2}
$$
A: Let $L=\lim_{x \rightarrow 0^+}\frac{1}{x^2}\int_{0}^{x}t^{1+t}dt = \lim_{x \rightarrow 0^+}\frac{\int_{0}^{x}t^{1+t}dt}{x^2}$.
Note that both the numerator and the denominator tend to $0$ as $x$ tends to $0$. It follows by l'Hopital's Rule that
$L=\lim_{x \rightarrow 0^+}\frac{\frac{d}{dx}\int_{0}^{x}t^{1+t}dt}{\frac{d}{dx}x^2} = \lim_{x \rightarrow 0^+} \frac{x^{1+x}}{2x}$.
Here we have made use of the fundamental theorem of calculus in the numerator. Carrying on, 
$L=\frac{1}{2} \lim_{x \rightarrow 0^+} x^{1+x}x^{-1}=\frac{1}{2} \lim_{x \rightarrow 0^+} x^{x} = \frac{1}{2}$.
A: Use the substitution $u=t^2$ to reduce the expression under limit to $$\frac{1}{2x^2}\int_{0}^{x^2}(\sqrt {u}) ^{\sqrt{u}} \, du$$ By Fundamental Theorem of Calculus the desired limit equals $1/2$ as the integrand above tends to $1$ as $u\to 0$.
