Completing the square to solve limit problems

Question

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!

Answer 1

It need not be $\alpha^2$. Adding any constant $\beta$ doesn't change the limit:

$$\lim\limits_{x\to\infty}\sqrt{x^2+2\alpha x+\beta}-\sqrt{x^2+2\alpha x}=\lim\limits_{x\to\infty}\dfrac{\beta}{\sqrt{x^2+2\alpha x+\beta}+\sqrt{x^2+2\alpha x}}= 0$$

Answer 2

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*}

Answer 3

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

Answer 4

More in general

$$\sqrt[n]{x^n+n\alpha x^{n-1}}-\sqrt[n]{(x+\alpha)^n}\to 0$$

indeed by binomial first order expansion

$$\sqrt[n]{x^n+n\alpha x^{n-1}}=x\left(1+\frac{n\alpha}{x}\right)^\frac1n=x\left(1+\frac{\alpha}{x}+o\left(\frac1x\right)\right)=x+\alpha+o(1)$$

therefore

$$\sqrt[n]{x^n+n\alpha x^{n-1}}-\sqrt[n]{(x+\alpha)^n}=x+\alpha+o(1)-(x+\alpha) =o(1)\to 0$$

Answer 5

If $f(x)\to \infty$ and $g(x)^2/f(x)\to 0$ as $x\to \infty$ then $g(x)/f(x)\to 0$ because $f(x)>0\implies|g(x)/f(x)|=\sqrt {g(x)^2/f(x)}\cdot 1/\sqrt {f(x)}.$

Then $\sqrt {f(x)+g(x)}-\sqrt {f(x)}\to 0$ because, when $x$ is large enough that $f(x)>0$ and $1+g(x)/f(x)>0,$ we have $$\left|\sqrt {f(x)+g(x)}-\sqrt {f(x)}\right|=\left|\sqrt {f(x)}\cdot\frac {g(x)/f(x)}{1+\sqrt {1+g(x)/f(x)}}\right |\le$$ $$\le \left|\sqrt {f(x)}\cdot\frac {g(x)}{f(x)}\right|=$$ $$=\sqrt {g(x)^2/f(x)}.$$

E.g. $\sqrt {x^5+7x^2+8}-\sqrt {x^5}\to 0$ as $x\to \infty.$

Answer 6

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}