$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} -n$? 
Calculate
  $\displaystyle\lim_{n \to \infty}
\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right)$.

$\displaystyle\lim_{n \to \infty}\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right) =
\infty - \infty$ We have an indeterminate form 
So I proceeded to factorize $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1   \right]$$
taking the limit:
$$\lim\limits_{n \rightarrow \infty} n \left[ \sqrt{\frac{n+1}{n}}-1   \right]= \infty \cdot 0$$
indeterminate again
What am i missing? How is the way forward to proceed? Much appreciated

Using you approach $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1   \right]=n \left[ \sqrt{1+\frac{1}{n}}-1   \right]$$ Now, use the generalized binomial theorem or Taylor series to get, for small values of $x$, $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ $$\sqrt{1+x}-1=\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Make now, since $n$ is large, $x=\frac 1n$ $$\sqrt{1+\frac{1}{n}}-1=\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$n \left[ \sqrt{1+\frac{1}{n}}-1   \right]=\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
Try with $n=10$ (which is very small, so you will have $\sqrt{110}-10\approx 0.488088$ while the above truncated series would give $\frac{39}{80}=0.487500$.
