If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$

HINT
Start with rearranging the proposed expression as follows:
\begin{align*}
\sqrt{(n + a)(n + b)} - n & = \frac{(n + a)(n + b) - n^{2}}{\sqrt{(n + a)(n + b)} + n}\\\\
& = \frac{(a + b)n + ab}{\sqrt{(n + a)(n + b)} + n}\\\\
& = \frac{(a + b) + ab/n}{\sqrt{(1 + a/n)(1 + b/n)} + 1}
\end{align*}
Now it remains to take the limit as $n$ approaches infinity.
Can you take it from here?
