Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$ 
Let $\{x_n\}_{n\geq 1}$ be defined as $x_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$. Then $\lim\limits_{n\to\infty} x_n$ is?

I want to find limit of this problem by a very specific method . I am uploading the pic of that method and my attempt.
Please guide me as to how to take this method ahead. 


Starting where you left: for $n\geq 1$,
$$
x_n = \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{k}{n^3}} \tag{1}
$$
While this furiously looks like a Riemann sum, it is not one due to the $\frac{k}{n^3}$ in the denominator. But then, we can use the squeeze theorem: as $1\leq k\leq n$,
$$
\frac{1}{1+\frac{1}{n^2}}\cdot \underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k}{n}}_{}
= \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{1}{n^2}}
\leq x_n \leq  \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{1}{n^3}}
= \frac{1}{1+\frac{1}{n^3}}\cdot\underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k}{n}}_{}
$$
and since $\lim_{n\to\infty}\frac{1}{1+\frac{1}{n^2}} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n^3}} = 1$, by the squeeze theorem the limit will be that of
$$
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{k}{n} = \int_0^1 f(x)dx
$$
for some very simple function $f$.
