Am I missing something in the answer to Spivak's Calculus, problem 5-39(vi)? 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 $x&gt;0$, $$x\left(\sqrt{x+2}-\sqrt{x}\right)=\sqrt{x}\left(\sqrt{x^{2}+2x}-x\right)$$ 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.
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$