# Completing the square to solve limit problems

There’s a trick I’ve been using to solve a common class of limit problems for a while now. I’ve never seen it taught in a textbook, but I once wrote out a few lines of work to justify it to myself in one of my notebooks. Here is a sample problem to illustrate my technique:

$$\lim_{x\to\infty}\sqrt{x^2+x}-x=\lim_{x\to\infty}\sqrt{x^2+x+\frac14}-x=\lim_{x\to\infty}\left(x+\frac12\right)-x=\frac12$$

It’s such a shortcut compared to rationalization or however you’re “supposed” to solve that, and I’m quite certain that it’s valid. But I’m starting to feel a little leery posting this as a solution to MSE problems since I don’t quite remember the few lines of justification all those years ago. Could someone please provide a proof that $$\lim_{x\to\infty}\sqrt{x^2+2\alpha x}-\sqrt{x^2+2\alpha x+\alpha^2}=0$$ or whatever equivalent formulation you would prefer? I’m sure that delta-epsilon drudgery is not necessary at all. (If nobody gets to this by the end of the day, I’ll self-answer just to have something to link to.)

Thanks!

It need not be $$\alpha^2$$. Adding any constant $$\beta$$ doesn’t change the limit:
$$\lim\limits_{x\to\infty}\sqrt{x^2+2\alpha x+\beta}-\sqrt{x^2+2\alpha x}=\lim\limits_{x\to\infty}\dfrac{\beta}{\sqrt{x^2+2\alpha x+\beta}+\sqrt{x^2+2\alpha x}}= 0$$