Prove $\lim_{n\to\infty}\sqrt{(n+a)(n+b)}-n=\frac{a+b}2$ Prove $$\lim_{n\to\infty}\sqrt{(n+a)(n+b)}-n=\frac{a+b}2.$$
I have found the preceding problem in the Calculus book (Bartle & Sherbert) where I am asked to demonstrate the previous limit assuming $a>0$, $b>0$. 
I have been trying for an hour, but I cannot see how I can demonstrate it.
 A: Hint. One may write, as $n \to \infty$,
\begin{align}
\sqrt{(n+a)(n+b)}-n&=\frac{\left(\sqrt{(n+a)(n+b)}-n\right)\left(\sqrt{(n+a)(n+b)}+n\right)}{\sqrt{(n+a)(n+b)}+n}\\\\
&=\frac{(n+a)(n+b)-n^2}{\sqrt{(n+a)(n+b)}+n}
\end{align}
Hope you can take it from here.
A: Remember that $$\sqrt{x}-\sqrt{y}= {x-y\over \sqrt{x}+\sqrt{y}}$$
so
$$\lim_{n \to \infty} \sqrt{(n+a)(n+b)} -n = \lim_{n \to \infty} {(n+a)(n+b) -n^2 \over \sqrt{(n+a)(n+b)} +n}$$
$$= \lim_{n \to \infty} {n\Big((a+b)+{ab\over n}\Big) \over n(\sqrt{(1+{a\over n})(1+{b\over n})} +1)}$$
$$=\frac{a+b}{2}$$
A: HINT
As an alternative by binomial expansion $$(1+x)^n=1+xn+o(x) \quad x\to 0$$
we have
$$\sqrt{(n+a)(n+b)} -n =n\sqrt{1+\frac a n}\sqrt{1+\frac b n} -n =$$$$= n\left(1+\frac a {2n}+o\left(\frac1n\right)\right)\left(1+\frac b {2n}+o\left(\frac1n\right)\right)-n=\ldots$$
A: You may consider this sequence limit in the context of the AGM inequality, relating the geometric and the arithmetic mean. The identity
$$\sqrt{XY}\;=\;\frac{X+Y}2\:-\:\frac12\left(\sqrt X - \sqrt Y\,\right)^2$$
is a "quantitative AGM version" since the last term involving the square measures the spread between the geometric and the arithmetic mean.
Now let $\,X=n+a, Y=n+b,\,$ and subtract $n$ to obtain
$$\sqrt{(n+a)(n+b)}-n\;=\;\frac{a+b}2
\:-\:\frac12\left(\sqrt{n+a}-\sqrt{n+b}\,\right)^2$$
The last term is obviously a null sequence, each element of which is negative (if $a\neq b$).
Thus the limit is approached from below.
