Am I missing something in the answer to Spivak's Calculus (4E), problem 5-39(vi)?

Earlier, problem 5-39(v)(p. 113) has established that $$\mathop{\lim}\limits_{x \to \infty}\sqrt{x^{2}+2x}-x=1$$ and then problem 5-39(vi) asks for $$\mathop{\lim}\limits_{x \to \infty}x\left(\sqrt{x+2}-\sqrt{x}\right).$$

Spivak's key (Combined Answer Book, p.78) determines this answer by going through several algebraic manipulations; but for $x>0$, $$x\left(\sqrt{x+2}-\sqrt{x}\right)=\sqrt{x}\left(\sqrt{x^{2}+2x}-x\right)$$ so that $$\mathop{\lim}\limits_{x \to \infty}x\left(\sqrt{x+2}-\sqrt{x}\right)=\left(\mathop{\lim}\limits_{x \to \infty}\sqrt{x}\right)\cdot1$$ can be determined by simply substituting the limit found in the preceding problem. Does taking this "shortcut" miss some required steps?

I wouldn't ask, except that where Spivak can refer to the answer to a previous problem to save work and space, he nearly always does. When he goes to the trouble of working out all the steps it is usually to illustrate something that would otherwise have been missed. But it seems that here the thing to observe is that the previously found limit exists and can be used.

  • 2
    $\begingroup$ Your proof is allright. The several algebraic manipulations that you mention and that are needed to get an alternative proof are not so complicated either. $\endgroup$
    – Did
    Sep 20, 2011 at 21:14
  • $\begingroup$ @Didier: yes, in fact there just about the same ones used for (v). My concern that I was missing some reason for repeating them explicitly, rather than using (as is really the emphasis of much of the chapter, and is precisely the point of (iv)) limits we've established, when we can. $\endgroup$
    – orome
    Sep 20, 2011 at 21:19

2 Answers 2


I see nothing wrong with your argument. The only possibly delicate point is that $\lim\limits_{x\to\infty}\sqrt{x}=\infty$. Since $\infty\cdot1$ is not an indeterminate form, everything seems copacetic.


$\lim_{x \to \infty}(\sqrt{x+2}-\sqrt{x})\frac{(\sqrt{x+2}+\sqrt{x})}{(\sqrt{x+2}+\sqrt{x})}x=$

$=\lim_{x \to \infty}\frac{x+2-x}{\sqrt{x+2}+\sqrt{x}}x=\lim_{x \to \infty}\frac{2x}{\sqrt{x+2}+\sqrt{x}}=$

$=\lim_{x \to \infty}\frac{2x/x}{\sqrt{x+2}/x+\sqrt{x}/x}$$=\lim_{x \to \infty}\dfrac{2}{\sqrt{\frac{1}{x}+\frac{2}{x^2}}+\sqrt{\frac{1}{x}}}=\frac{2}{0}=\infty$

  • $\begingroup$ Thanks, but that's not what the question was about. That much is in the answer key. The question was why the key didn't do the substitution above instead (and if there was a reason, whether it was important conceptually). $\endgroup$
    – orome
    Sep 21, 2011 at 3:41
  • $\begingroup$ why $\frac20=\infty$? That's the answer in the solutions manual too but it's not explained. $\endgroup$
    – super.t
    Nov 14, 2020 at 18:35

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