<s>[INST] Epsilon-delta proof of $\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2}$ I have been doing $\varepsilon$-$\delta$ proofs for fun and I challenged myself to prove $$\displaystyle\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2},\quad a,b\in\mathbb{R}$$
The definition says: We say that $\displaystyle\lim_{x\to\infty}f(x)=l$ if for any positive number $\varepsilon$ we can find a positive number $N$ (depending on $\varepsilon$ in general) such that $|f(x)-l|<\varepsilon$ whenever $x>N$.
So I started with: $\left|\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}\right|<\varepsilon$ whenever $x>N$.
Manipulating the first inequatlity
\begin{gather*}
-\varepsilon<\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}<\varepsilon\\
-\varepsilon+\dfrac{a+b}{2}<\sqrt{(x+a)(x+b)}-x<\varepsilon+\frac{a+b}{2}
\end{gather*}
At this point I thought about adding $x$, squaring the expressions and then expanding them. I did it and I got: $$\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2-a\varepsilon-b\varepsilon-2x\varepsilon+\varepsilon^2<x^2+ax+bx+ab<\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2+a\varepsilon+b\varepsilon+2x\varepsilon+\varepsilon^2$$
And here I'm not sure how to proceed. Am I on the right track? 
Thanks for any help / hints. 
Note: There might be other ways to prove this, but I'd like to do it using just algebra if possible, even if it's not the best method. 
 [/INST] $$\lim_{x\to\infty}\sqrt{(x+a)(x+b)}-x-\frac{a+b}2\\
=\lim_{x\to\infty}\frac{(x+a)(x+b)-(x+\frac{a+b}2)^2}{\sqrt{(x+a)(x+b)}+x+\frac{a+b}2}$$
</s>