# 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?

• Don't be annoyed that you can't integrate $(1+u)^{1/u}.$ Nobody can. Jun 30, 2019 at 9:33

$$(1+u)^{1/u}=e^{\frac 1 u \log(1+u) }\leq e$$. 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.

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 $$e=f(0)=F'(0)=$$ $$=\lim_{x\to 0} \frac {F(x)-F(0)}{x-0}=$$ $$=\lim_{x\to 0}\frac {F(x)}{x}.$$

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 $$x→ 0.$$

• The function has to be continuous in the closed interval. Jun 29, 2019 at 23:35
• It can be extended to continuous on $[0,x]$ since singularity at $0$ is removable-- integrand goes to $e$ as $x \to 0^+.$ Jun 29, 2019 at 23:45
• @DiegoMath I forgot to notify you in last comment. Could you read the revised post and see if you agree it works as answer? Thanks. [not seeking points...] Jun 30, 2019 at 12:07