TITLE: Completing the square to solve limit problems
QUESTION [47 upvotes]: 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!

REPLY [2 votes]: Let $u=x+\alpha$. Then, $u\to\infty$ as $x\to\infty$ and
\begin{align}\lim_{x\to\infty}\sqrt{x^2+2\alpha x}-\sqrt{x^2+2\alpha x+\alpha^2}&=\lim_{x\to\infty}\sqrt{(x+\alpha)^2-\alpha^2}-\sqrt{(x+\alpha)^2}\\&=
\lim_{u\to\infty}\sqrt{u^2-\alpha^2}-\sqrt{u^2}\\&=
\lim_{u\to\infty}|u|\left(\sqrt{1-\frac{\alpha^2}{u^2}}-1\right)\\&=
\lim_{u\to\infty}|u|\frac{\left(\sqrt{1-\frac{\alpha^2}{u^2}}-1\right)\left(\sqrt{1-\frac{\alpha^2}{u^2}}+1\right)}{\sqrt{1-\frac{\alpha^2}{u^2}}+1}\\&=
\lim_{u\to\infty}-\frac{\alpha^2}{u^2}\frac{|u|}{\sqrt{1-\frac{\alpha^2}{u^2}}+1}\\&=
0
\end{align}