Epsilon-delta proof of $\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2}$

Question

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.

Answer 1

The "rationalization" trick are the same, but this may be the rigorous $\varepsilon$-$\delta$ proof you want.

Write $$\sqrt{(x + a)(x + b)} - x - \frac{a + b}{2} = \frac{(x + a)(x + b) - \left(x + \frac{a + b}{2}\right)^2}{\sqrt{(x + a)(x + b)} + \left(x + \frac{a + b}{2}\right)} = \frac{-\frac{(a - b)^2}{4}}{\sqrt{(x + a)(x + b)} + \left(x + \frac{a + b}{2}\right)}.$$

Therefore, $$ \left|\sqrt{(x + a)(x + b)} - x - \frac{a + b}{2}\right| = \frac{C}{\sqrt{(x + a)(x + b)} + \left(x + \frac{a + b}{2}\right)} < \frac{C}{x + \frac{a + b}{2}} \tag{1} $$ where $C = \dfrac{(a - b)^2}{4} \geq 0$, note the denominator is positive when $x$ is sufficiently large (say, $x > -(a + b)/2$) so the absolute value symbol can be removed. Given $\varepsilon > 0$, take $\delta = \max\left(\dfrac{C}{\varepsilon} - \dfrac{a + b}{2}, 1\right) > 0$, it then follows by $(1)$ that for all $x > \delta$, $$\left|\sqrt{(x + a)(x + b)} - x - \frac{a + b}{2}\right| < \varepsilon.$$

Answer 2

This is not an epsilon delta proof, but here it goes:

$$\left(\sqrt{(x+a)(x+b)}-x\right)= x(\sqrt{(1+\frac{a}{x})(1+\frac{b}{x})}-1) = x(f(x) -1)$$

expanding $g(z) := f(\frac{1}{x})$ around $z=0$: $g(z) = 1 + \frac{a+b}{2}z +\mathcal{O}(z^2)$ therfore

$$ x(f(x) -1) = x(1 + \frac{a+b}{2}\frac{1}{x} + \mathcal{O}(\frac{1}{x^2}) -1) = \frac{a+b}{2} + \mathcal{O}(\frac{1}{x}) \overset{x\rightarrow \infty}{\longrightarrow} \frac{a+b}{2} $$

Ok just now I read your note. Please don't down vote because I'm unable to read posts till the end :)

Answer 3

$$\lim_{x\to\infty}\sqrt{(x+a)(x+b)}-x-\frac{a+b}2\\ =\lim_{x\to\infty}\frac{(x+a)(x+b)-(x+\frac{a+b}2)^2}{\sqrt{(x+a)(x+b)}+x+\frac{a+b}2}$$

Answer 4

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

$$ \lim_{x \rightarrow \infty} \left( \sqrt{ ( x + a ) ( x + b ) } - x \right) = \frac{a+b}{2}. $$