Integral as a limit of sum: $\lim\limits_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)$ 
The value of$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)=\,?$$

Can the limit be partially applied to the denominator after converting the numerator into an integral?
I wrote this as $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\left(\frac {k/n}{1+1/n+2k/n^2}\right)$$
From what I know I can write that $1/n$ as $\mathrm{d}x$ and $k/n$ as $x$. Can I apply limit to denominator as $n$ tends to infinity and rewrite the denominator as $1$.

One the one hand
$$
\mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2  + n + 2k}}}  \le \mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 }}}  = \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}}  = \int_0^1 {x\, dx}  = \frac{1}{2}.
$$
On the other hand
\begin{align*}
\sum\limits_{k = 1}^n {\frac{k}{{n^2  + n + 2k}}}  \ge \sum\limits_{k = 1}^n {\frac{k}{{n^2  + 3n}}} & = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{1 + \frac{3}{n}}}}  \ge \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\left( {1 - \frac{3}{n}} \right)} \\ & = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}}  - \frac{3}{n}\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} ,
\end{align*}
i.e.,
\begin{align*}
\mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2  + n + 2k}}} & \ge \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}}  - \mathop {\lim }\limits_{n \to  + \infty } \frac{3}{n}\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} \\ & = \int_0^1 {x\, dx}  - 0 \cdot \int_0^1 {x\, dx}  = \frac{1}{2}.
\end{align*}
Thus, by the squeeze theorem, the limit is $\frac{1}{2}$.
