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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!

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    $\begingroup$ How did you go from $x^2 + x$ to $x^2 + x + \frac{1}{4}$? $\endgroup$ – Matti P. Nov 7 at 11:21
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    $\begingroup$ That's the proof he wants us work on :P Rationalizing should do $\endgroup$ – pooja Nov 7 at 11:23
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    $\begingroup$ What about trying the rationalization for the second limit? Sorry but written that way, I read it as asking people to do the work for you. -1 $\endgroup$ – Taladris Nov 8 at 13:27
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    $\begingroup$ @Taladris You're not wrong. I was answering a different question math.stackexchange.com/questions/3425638/… and wanted a quick justification to link to more than I wanted the extra rep from self-answering later in the day. I'm grateful to pooja and tracelocation for answering my question as quickly as they did despite the lack of context. $\endgroup$ – Matthew Daly Nov 8 at 14:03
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    $\begingroup$ I'm sorry to be that person but I don't think you have a shortcut here. I think it seems that way because you actually have an unclear proof. You can see from the answers... i.e. the bit where you do the 'shortcut' is where you leave out the details of the proof, which typically involve rationalization or taylor expansion or something you are trying to avoid. $\endgroup$ – T_M Nov 9 at 8:58
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It need not be $\alpha^2$. Adding any constant $\beta$ doesn't change the limit:

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

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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*}

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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

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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

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

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  • $\begingroup$ More general, the limit of the n-th root of $x^n + nax^{n-1} + o(x^{n-1})$ is x+a. $\endgroup$ – gnasher729 Nov 10 at 0:48
  • $\begingroup$ More precisely $x+a+o(1)$. $\endgroup$ – user Nov 10 at 0:51
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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 $x\to \infty.$

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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}

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