<s>[INST] 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
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
 [/INST] Another fun solution involves the HM-GM-AM-Inequality:
By the mentioned inequality, we obtain
$$
\frac{2}{\frac1{n+a} + \frac1{n+b}} \leq \sqrt{(n+a)(n+b)} \leq \frac{n+a+n+b}2
$$
Now note that
$$
\frac{2}{\frac1{n+a} + \frac1{n+b}} = \frac{2(n+a)(n+b)}{n+a+n+b} = \frac{n (2n + a + b) + n(a+b)+ab}{2n+a+b} \geq n + \frac{n(a+b)}{2n+a+b}.
$$
By L'Hospital's rule, we have $\lim_{n \rightarrow \infty}\frac{n(a+b)}{2n+a+b} = \frac{a+b}2$. Letting $n \rightarrow \infty$ and subtracting $n$
in the original inequality, this yields
$$
\frac{a+b}2 \leq \lim_{n \rightarrow \infty} \left( \sqrt{(n+a)(n+b)}-n \right) \leq \frac{a+b}2
$$
and we are done.
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