<s>[INST] $\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} -n$? 
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
 [/INST] Trick is to write $\sqrt{n^2 + n} - n$ in a nice form.Below we use rationalization
$\lim _{n\rightarrow \infty}\sqrt{n^2 + n} - n = \lim_{n \rightarrow \infty} \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n \rightarrow \infty} \sqrt{n}(\frac{1}{\sqrt{n+1}+\sqrt{n}}) = \lim_{n \rightarrow \infty} \frac{\sqrt{n}(1)}{\sqrt{n}(\sqrt{1+\frac{1}{n}} + 1)} = \frac{1}{2}$
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