<s>[INST] Simplify _Elementary Calculus_ section 1.6 problem 25 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!
 [/INST] From this step:
$$=\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})}$$
you have 
$$=\frac{-(c-7)}{(c-7)(3+\sqrt{c+2})}$$
Cancel the common factor and the rest should follow straightforwardly.
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