= Solution
Use <rationalization of a square-root difference> to avoid subtracting two large, nearly equal quantities:
$$
\sqrt{n^2+n}-n
=\frac{n}{\sqrt{n^2+n}+n}
=\frac1{\sqrt{1+1/n}+1}.
$$
Since $1/n\to0$ and the <square root> is continuous at $1$, the denominator tends to $2$. The reciprocal limit law established above is applicable because this limit is nonzero. Hence
$$
\boxed{\lim_{n\to\infty}\bigl(\sqrt{n^2+n}-n\bigr)=\frac12.}
$$
\b[The limit is one half; rationalization exposes the finite difference hidden by cancellation.]
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