Computing $\lim_{A\to\infty} \frac{1}{A} \int\limits_1^A \! A^{\frac{1}{x}} \, \mathrm{d}x.$

Squeezing seems to provide a quick solution. Assume $A>e^2$. We have:

$$ A^{\frac{1}{x}}=\exp\left(\frac{\log A}{x}\right)\geq 1+\frac{\log A}{x}\tag{1}$$ hence: $$ \int_{1}^{A}A^{\frac{1}{x}}\,dx \geq A+\left(\log^2(A)-1\right).\tag{2} $$ On the other hand: $$\begin{eqnarray*}\int_{1}^{A}A^{\frac{1}{x}}\,dx = \int_{\frac{1}{A}}^{1}\frac{A^x}{x^2}\,dx &=&(A-1)-\left.\frac{A^x-1}{x}\right|_{\frac{1}{A}}^{1}+\log(A)\int_{\frac{1}{A}}^{1}\frac{A^x}{x}\,dx\\&=&A\left(A^{\frac{1}{A}}-1\right)+\log^2(A)+\log(A)\int_{\frac{1}{A}}^{1}\frac{A^x-1}{x}\,dx \end{eqnarray*}$$ and $\frac{A^x-1}{x}$ is a convex function on $(0,1]$, hence it is not difficult to find a tight upper bound for the LHS of $(2)$ through the Hermite-Hadamard inequality or Jensen's inequality. Another chance is given by proving, then exploiting, $$ \forall x\in(0,1),\qquad \frac{A^{1-x}-1}{1-x}\leq (A-1)\exp\left[\log\left(\frac{\log A}{A-1}\right)\,x\right]\tag{3} $$ so by $(2)$ and $(3)$ it follows that the wanted limit is just $\color{red}{1}$.

$$\begin{aligned}\lim_{A\to+\infty}\frac1A\int_1^AA^{\frac1x}\,dx&\stackrel{\ln A=xt}{\!=\!=\!=}\lim_{A\to+\infty}\frac{\int_{1}^{\ln A}t^{-2}e^{t}\,dt-\int_{1}^{\frac{\ln A}{A}}t^{-2}e^{t}\,dt}{\frac{A}{\ln A}} \notag \\&\stackrel{(*)}{=}\lim_{A\to+\infty}\frac{d\left(\int_{1}^{\ln A}t^{-2}e^{t}\,dt-\int_{1}^{\frac{\ln A}{A}}t^{-2}e^{t}\,dt\right)/dA}{d\left(\frac{A}{\ln A}\right)/dA}\notag \\ &=\lim_{A\to+\infty}\frac{1+A^{1/A}(\ln A-1)}{\ln A-1}=1\end{aligned}$$

$(*)$ using strong DLH.