Propose by Jan Šustek

We can use L'Hopital's rule to evaluate this:

${ \lim }_{ a\to \infty }\frac { 1 }{ a } \int _{ 1 }^{ a } a^{ \frac { 1 }{ x } }\, dx\quad =\quad \lim _{ a\rightarrow \infty }{ \frac { d }{ da } \int _{ 1 }^{ a } a^{ \frac { 1 }{ x } }\, dx } \quad =\quad \lim _{ a\rightarrow \infty }{ \frac { d }{ da } \left[ F(a)-F(1) \right] }$

where $F(x)$ is the antiderivative of $a^{ \frac { 1 }{ x } }$

$\lim _{ a\rightarrow \infty }{ \frac { d }{ da } \left[ F(a)-F(1) \right] } =\quad \lim _{ a\rightarrow \infty }{ F'(a) } =\quad \lim _{ a\rightarrow \infty }{ { a }^{ \frac { 1 }{ a } } }$

because $F'(x)$ is ${ a }^{ \frac { 1 }{ x } }$ and thus $F'(a)$ is ${ a }^{ \frac { 1 }{ a } }$ .

This last limit can be evaluated by using exponents, and then again with L'hopital's rule:

$\lim _{ a\rightarrow \infty }{ { a }^{ \frac { 1 }{ a } } } =\quad { e }^{ \lim _{ a\rightarrow \infty }{ \frac { 1 }{ a } \ln { a } } }=\quad { e }^{ \lim _{ a\rightarrow \infty }{ \frac { \ln { a } }{ a } } }\quad =\quad { e }^{ \lim _{ a\rightarrow \infty }{ \frac { 1 }{ a } } }=\quad { e }^{ 0 }\quad =\quad 1$