Neither \sqrt{n^2+n} nor n converges so it's necessary to transform the expression to the form on which we can apply algebra of limits.

\sqrt{n^2+n}-n

=(\sqrt{n^2+n}-n)\cdotp \dfrac{\sqrt{n^2+n}+n }{\sqrt{n^2+n} +n}

=\dfrac{(\sqrt{n^2+n})^2-n^2}{\sqrt{n^2+n} +n} \\Using the difference of squares formula

=\dfrac{n}{\sqrt{n^2+n} +n}

=\dfrac{1}{\sqrt{1+\dfrac{1}{n} }+1 } \\divide through by n

Therefore

\lim\limits_{n\to \infty} \sqrt{n^2+n}-n

= \lim\limits_{n\to \infty} \dfrac{1}{\sqrt{1+\dfrac{1}{n} }+1 }

=\dfrac{1}{2}