Step 1: Recognise a limit of a sum.
We must find $\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{r=1}^n r\,e^{r/n}$. Sums like this, with $\tfrac{1}{n}$ and $\tfrac{r}{n}$, usually turn into a definite integral.
Step 2: Rearrange the sum.
Pull one $\tfrac{1}{n}$ outside and keep one $\tfrac{1}{n}$ inside: \[ \frac{1}{n}\sum_{r=1}^n\left(\frac{r}{n}\right)e^{r/n}. \]
Step 3: Convert to an integral.
As $n\to\infty$, replace $\tfrac{r}{n}$ by $x$ and $\tfrac{1}{n}$ by $dx$, with $x$ running from $0$ to $1$: \[ L=\int_0^1 x\,e^x\,dx. \]
Step 4: Integrate by parts.
Take $u=x$ and $dv=e^x\,dx$, so $du=dx$ and $v=e^x$. Then \[ \int x\,e^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=e^x(x-1). \]
Step 5: Apply the limits $0$ to $1$.
\[ \big[e^x(x-1)\big]_0^1=e^1(1-1)-e^0(0-1)=0-(-1)=1. \]
Step 6: State the answer.
So the limit equals $1$.
\[ \boxed{1} \]