Find the value of the following (using an epsilon proof or basic limit properties (no L'hospital)):
$$\lim_{x\to\infty}\left( \sqrt{(x+a)(x+b)}-x \right)\forall a,b\in\mathbb{R}$$
I've tried rewriting it in several ways, but I don't seem to bet getting very far; I always end up with something in indeterminate form. How can you prove the value of this? Any hints?
On
Hint: $$\sqrt{(x+a)(x+b)}=\sqrt{x^2+(a+b)x+ab}=\sqrt{\left(x+\frac{a+b}2\right)^2+ab-\frac{(a+b)^2}4}$$
Bonus:
Can you generalize this for $n$ constants $a_1,a_2,\ldots,a_n$ with the surd being $\sqrt[n]{\cdot}$ ?
Hint for the generalization:
$$(x+y)^n=x^n+nyx^{n-1}+\ldots$$
What is the coefficient of $x^{n-1}$ in $\prod\limits_1^n (x-a_i)$ scaled by $1/n$ ?
\begin{align*} \lim_{x\to\infty}\left( \sqrt{(x+a)(x+b)}-x \right)&=\lim_{x\to\infty}\frac{(x+a)(x+b)-x^2}{ \sqrt{(x+a)(x+b)}+x }\\ &=\lim_{x\to\infty}\frac{(a+b)x+ab}{ \sqrt{(x+a)(x+b)}+x }\\ &=\lim_{x\to\infty}\frac{(a+b)+\frac{ab}{x}}{ \sqrt{(1+\frac{a}{x})(1+\frac{b}{x})}+1 }\\ &=\frac{a+b}{2} \end{align*}