Compute $\lim \limits_{n\to \infty} \frac{1}{n}\sum_{i,j=1}^n\frac{1}{\sqrt{i^2+j^2}}$

Denote the limit you have to compute by $L$.
We may apply the Stolz-Cesaro theorem to get that $L=\lim\limits_{n\to \infty}\left(2\sum_{i=1}^{n-1}\frac{1}{\sqrt{n^2+i^2}}+\frac{1}{\sqrt{2n^2}}\right)$.
Since $\lim\limits_{n\to \infty}\frac{1}{\sqrt{2n^2}}=0$ and $\lim\limits_{n\to \infty}\sum_{i=1}^{n-1}\frac{1}{\sqrt{n^2+i^2}}=\int\limits_0^1\frac{dx}{\sqrt{x^2+1}}=\ln(1+\sqrt 2)$, we get that $L=2\ln(1+\sqrt 2)$.

Define $$u_n=\sum_{\substack{1 \leq i,j \leq n \\ i=n \text { or } j=n}}{(i^2+j^2)^{-1/2}}.$$

You want to find $\lim \, \frac{1}{n}\sum_{k=1}^n{u_k}$.

It is, provided that $\{u_n\}$ converges, the limit of $\{u_n\}$.

Now, $u_n=2v_n-(\sqrt{2}n)^{-1}$ where $v_n=\sum_{k=1}^n{\frac{1}{\sqrt{n^2+k^2}}}$.

Now, $v_n=\frac{1}{n} \sum_{k=1}^n{(1+(k/n)^2)^{-1/2}} \rightarrow \int_0^1{\frac{dx}{\sqrt{x^2+1}}}=\operatorname{arsinh}(1)$.

Thus your limit is $2\operatorname{arsinh}(1)$.