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
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$

Just take the difference
of the expression and its
proposed limit
and conjugate that.
Works out very nicely.
$\begin{array}\\
d(n)
&=\sqrt{(n+a)(n+b)}-(n+\frac{a+b}{2})\\
&=(\sqrt{(n+a)(n+b)}-(n+\frac{a+b}{2}))\dfrac{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{(n+a)(n+b)-(n+\frac{a+b}{2})^2}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{n^2+(a+b)n+ab-(n^2+n(a+b)+\frac{(a+b)^2}{4})}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{ab-\frac{(a+b)^2}{4}}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{-\frac{(a-b)^2}{4}}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{-(a-b)^2}{4\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2}))}\\
\text{so}\\
|d(n)|
&=|\dfrac{(a-b)^2}{4(\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2}))}|\\
&\lt|\dfrac{(a-b)^2}{4(2n)}|\\
&=|\dfrac{(a-b)^2}{8n}|\\
&\to 0\\
\end{array}
$
