How to compute: $ \lim\limits_{a\to \infty}\frac1a\int_1^a a^{\frac1x} dx$ Help me to compute the following limit 
$$ 
\lim_{a\to \infty}\frac1a\int_1^a a^{\frac1x} dx$$

I failed to apply the mean value theorem. And this the feature i had in mind when I saw this first

Any Hint?
 A: For $a>1$, by letting $t=\ln(a)/x$, then $dx=-\ln(a)dt/t^2$ and
$$\frac1a\int_1^a a^{\frac1x} dx=\frac{\ln(a)}{a}\int_{\ln(a)/a}^{\ln(a)} \frac{e^{t}}{t^2} dt=\frac{F(\ln(a))-F(\ln(a)/a)}{a/\ln(a)}$$
where $F(s)=\int_{1}^se^{t}/t^2 dt$.
Then by L'Hôpital's rule (note that $a/\ln(a)\to +\infty$ as $a\to+\infty$), it suffices to evaluate the limit
$$\frac{F'(\ln(a))(\ln(a))'-F'(\ln(a)/a)(\ln(a)/a)'}{(a/\ln(a))'}
=\frac{e^{\ln(a)/a}(\ln(a)-1)+1}{\ln(a)-1}
\to 1.
$$
A: Under $x=(a-1)t+1$, one has
\begin{eqnarray}
&&\frac1a\int_1^a a^{\frac1x} dx\\
&=&\frac{a-1}a\int_0^1 a^{\frac1{(a-1)t+1}} dt\\
&=&\frac{a-1}a\int_0^1 \exp\left({\frac{\ln a}{(a-1)t+1}}\right) dt.
\end{eqnarray}
Since, for large $a>0$,
\begin{eqnarray}
1&\le&\int_0^1 \exp\left({\frac{\ln a}{(a-1)t+1}}\right) dt\\
&=&\int_0^1 \sum_{n=0}^\infty\frac1{n!}\left({\frac{\ln a}{(a-1)t+1}}\right)^n dt\\
&=&1+\int_0^1\frac{\ln a}{(a-1)t+1}dt+\sum_{n=2}^\infty\frac{(\ln a)^n}{(a-1)^nn!}\int_0^1\left({\frac{1}{t+\frac{1}{a-1}}}\right)^n dt\\
&=&1+\frac{\ln a}{a-1}+\sum_{n=2}^\infty\frac{(\ln a)^n(1-a^{-n+1})}{(a-1)(n-1)n!}\\
&\le&1+\frac{\ln a}{a-1}+\sum_{n=2}^\infty\frac{(\ln a)^n}{(a-1)(n-1)n!}\\
&\le&1+\frac{\ln a}{a-1}+\frac{a}{(a-1)\ln a}
\end{eqnarray}
one has
\begin{eqnarray}
\lim_{a\to \infty}\int_0^1 \exp\left({\frac{\ln a}{(a-1)t+1}}\right) dt=1
\end{eqnarray}
or
$$ \lim_{a\to \infty}\frac1a\int_1^a a^{\frac1x} dx=\lim_{a\to \infty}\frac{a-1}{a}\int_0^1 \exp\left({\frac{\ln a}{(a-1)t+1}}\right) dt=1.$$
A: Consider:
$$\int_1^a a^{1/x}\, dx = \int_1^a \exp (x^{-1}\ln a )\, dx $$ $$= \int_1^a (\sum_{n=0}^{\infty}(x^{-1}\ln a)^n/n!\,)\, dx = \sum_{n=0}^{\infty} (\ln a)^n/n! \int_1^a x^{-n}\,dx$$ $$ = a-1 + (\ln a)^2 +\sum_{n=2}^{\infty} (\ln a)^n/n! \int_1^a x^{-n}\,dx.$$
Now
$$\sum_{n=2}^{\infty} (\ln a)^n/n! \int_1^a x^{-n}\,dx < \sum_{n=2}^{\infty} (\ln a)^n/n! \int_1^\infty x^{-n}\,dx $$ $$\tag 1= \sum_{n=2}^{\infty} \frac{1}{n-1}\frac {(\ln a)^n}{n!}.$$
Prove yourself a little lemma: For $x>0, \sum_{n=2}^{\infty} \dfrac{1}{n-1}\dfrac {x^n}{n!} \le 3\dfrac{e^x}{x}.$ It follows that $(1)$ is bounded above by $3\dfrac{a}{\ln a}.$
Summarizing, we have
$$a-1 \le \int_1^a a^{1/x}\, dx \le a-1 + (\ln a)^2 + 3\dfrac{a}{\ln a}.$$
Dividing by $a
$ shows the desired limit is $1.$
