Q:

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

A:

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

A:

Replacing $x=\frac{1}{t}$ and considering $t\to 0^+$ you get
\begin{eqnarray*} \sqrt{x^2+2\alpha x}-\sqrt{x^2+2\alpha x+\alpha^2}
& \stackrel{x=\frac{1}{t}}{=} & \frac{\sqrt{1+2\alpha t} - \sqrt{1+2\alpha t + a^2t^2}}{t} \\
& = & \frac{\sqrt{1+2\alpha t} - 1}{t} - \frac{\sqrt{1+2\alpha t + a^2t^2}-1}{t}\\
& \stackrel{t \to 0^+,L'Hosp.}{\longrightarrow}& \frac{\alpha}{\sqrt{1+2\alpha t}} - \frac{\alpha + t\alpha^2}{\sqrt{1+2\alpha t + a^2t^2}} \\
& = & \alpha - \alpha = 0
\end{eqnarray*}

A:

You can use $\sqrt{x^2+x}=x\sqrt {1+\frac 1x}=x(1+\frac 1{2x}+o(\frac 1x)) \to x+\frac 12+o(1)$ to justify what you do

