Calculating integral without L'hopital's rule

Problem$$\lim_{x \to 0}\frac{1}{x} \int_0^{x}(1+u)^{\frac{1}{u}}du=?$$

My first solution is using L'hopital's rule

Solution using L'hopital's rule, given limit
$$\lim_{x \to 0}\frac{(1+x)^{\frac{1}{x}}}{1}=e$$

I want another solution without using l'hopital's rule, but I can't found it. More annoying thing is I can't even integrate $$\int_0^{x}(1+u)^{\frac{1}{u}}du$$. Or It can't be solved without l'hopital's rule?

Answer

$(1+u)^{1/u}=e^{\frac 1 u \log(1+u) }\leq e$. Also $\log (1+u)=u+o(u)$ so, given $\epsilon >0$ there exists $\delta$ such that $\log(1+u) > (1-\epsilon) u$ for $0. Hence $e^{(1-\epsilon)} \leq(1+u)^{1/u} \leq e$ for $0 provided $0. Squeeze theorem completes the proof since $\epsilon$ is arbitrary.