Let's try this again. We're still on problem 25 in section 1.6 of Elementary Calculus.
$\frac{3-\sqrt{c+2}}{c-7}$
My first thought is (again) to multiply by $3+\sqrt{c+2}$:
$=\frac{(3-\sqrt{c+2})(3+\sqrt{c+2})}{(c-7)(3+\sqrt{c+2})}$ $=\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})}$ $=\frac{9-(c+2)}{3c+c\sqrt{c+2}-7\sqrt{c+2}-21}$
This looks "simplified" to me, so I proceed to substitute $c=7+\epsilon, \epsilon \in \mathbb{R}^*, \epsilon \approx 0$:
$=\frac{9-(9+\epsilon)}{3(7+\epsilon)+(7+\epsilon)\sqrt{7+\epsilon+2}-7\sqrt{9+\epsilon}-21}$ $=\frac{-\epsilon}{21+3\epsilon+7\sqrt{9+\epsilon}+\epsilon\sqrt{9+\epsilon}-7\sqrt{9+\epsilon}-21}$ $=-\frac{\epsilon}{3\epsilon + \epsilon\sqrt{9+\epsilon}}$
Knowing the answer is $-\frac{1}{6}$, it seems likely that this somehow reduces to $-\frac{\epsilon}{6\epsilon}$ (apart from some error), but I don't see how to get from $3\epsilon+\epsilon\sqrt{9+\epsilon}$ to $3\epsilon+3\epsilon=6\epsilon$.
Thanks for your help again!