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?
 A: $(1+u)^{1/u}=e^{\frac 1 u \log(1+u) }\leq e$. Also $\log (1+u)=u+o(u)$ so, given $\epsilon >0$ there exists  $\delta$ such that $\log(1+u) > (1-\epsilon) u$ for $0<u<\delta$. Hence  $e^{(1-\epsilon)} \leq(1+u)^{1/u} \leq e$ for $0<u<x$ provided $0<x<\delta$. Squeeze theorem completes the proof since $\epsilon$ is arbitrary. 
A: For $0<|u|<1$ let $$f(u)=(1+u)^{1/u}.$$ Then $f$ is continuous on $(-1,0)\cup (0,1).$ And $\lim_{u\to 0}f(u)=e.$ So let $f(0)=e.$
Now $f$ is continuous on $(-1,1)$.
For $x\in (-1,1)$ let $$F(x)=\int_0^xf(u)du.$$ By the Fundamental Theorem of Calculus, the continuity of $f$ implies that  $F'(x)=f(x)$ for all $x\in  (-1,1).$ Therefore $$e=f(0)=F'(0)=$$ $$=\lim_{x\to 0} \frac {F(x)-F(0)}{x-0}=$$ $$=\lim_{x\to 0}\frac {F(x)}{x}.$$
A: By the mean value theorem for integrals, the integral in your limit (with the $1/x$ in front but not with the limit) is same as $(1+c)^{1/c}$ for some $c \in [0,x].$ And as $x \to 0$ it forces $c \to 0.$
Correction: as noted by DiegoMath integrand needs to be continuous on $[0,x].$ However It can be extended to continuous on 
$[0,x]$
since singularity at 
$0$
is removable-- integrand goes to 
e
as 
$x→
0.$
