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
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
 A: 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?
A: Newton's Generalised Binomial expansion with $\ x=n^2,\ y = (a+b)n + ab,\ r = \frac{1}{2},\ $ works well.
A: 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.
A: 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}
$
