<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] The problem is that the $\epsilon$-method requires the result of the limit...
A "look-alike" method without knowing the result of the limit...
Given
$$
\lim_{x \rightarrow \infty} \left( \sqrt{ ( x + a ) ( x + b ) } - x \right).
$$
Let
$$
\sqrt{ ( x + a ) ( x + b ) } - x = \ell.
$$
Rewrite this as
$$
( x + a ) ( x + b ) = ( x + \ell)^2.
$$
So
$$
x^2 + ( a + b ) x + a b  = x^2 + 2 \ell x + \ell^2,
$$
or
$$
( a + b - 2 \ell ) x = \ell^2 - a b.
$$
So we can find $x$ for a given $\ell$, using
$$
x = \frac{ \ell^2 - a b }{ a + b - 2 \ell },
$$
and for the limit, we find $x \rightarrow \infty$.
We clearly see that
$$
\lim_{\ell \rightarrow \frac{a+b}{2}} x
= \lim_{\ell \rightarrow \frac{a+b}{2}} \frac{ \ell^2 - a b }{ a + b - 2 \ell } = \infty,
$$
consequently, we obtain
$$
\lim_{x \rightarrow \infty} \left( \sqrt{ ( x + a ) ( x + b ) } - x \right) = \frac{a+b}{2}.
$$
</s>