calculus - $limlimits_{n rightarrow infty} sqrt{n^2+n} -n$?

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

Answer

Hint: use the so-to-speak "multiply and divide by the conjugate" trick — it often helps to rationalize. In this case, since you're given a difference $\sqrt{n^2+n}-n$, multiply and divide by the sum of the same two terms $\sqrt{n^2+n}+n$:

$$\lim_{n\to\infty} \left(\sqrt{n^2+n}-n\right)=\lim_{n\to\infty} \frac{\left(\sqrt{n^2+n}-n\right)\left(\sqrt{n^2+n}+n\right)}{\sqrt{n^2+n}+n}=\cdots$$