Simplify _Elementary Calculus_ section 1.6 problem 25 (incorrect transcription) Given $c\neq7$ and $st(c)=7$, simplify $$\frac{3-\sqrt{c+2}}{\sqrt{c-7}}$$
My inclination, based on one of the examples in the book, is to multiply by $3+\sqrt{c+2}$, yielding:
$$\frac{(3-\sqrt{c+2})\cdot(3+\sqrt{c+2})}{\sqrt{c-7}\cdot(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{3\sqrt{c-7}+\sqrt{c-7}\sqrt{c+2}}$$
Substituting in $c=7+\epsilon$ leads to:
$$=\frac{9-(9+\epsilon)}{3\sqrt{\epsilon}+\sqrt{\epsilon}\sqrt{9+\epsilon}}$$
To get the answer in the back of the book, $-\frac{1}{6}$, I think one must somehow convert this into:
$$=-\frac{\epsilon}{6\epsilon}$$
I don't see a road from the simplification after the substitution to this, which makes me think I've made more fundamental errors. I specifically don't see a way to get from $\sqrt{\epsilon}\sqrt{9+\epsilon}$ to $3\sqrt{\epsilon}$, which seems necessary if I haven't made errors earlier.
Thanks for your help!
Edit: I have transcribed the problem wrong (I originally wrote $\sqrt{7-c}$ instead of $\sqrt{c-7}$) which explains the spurious absolute value in Andre's answer, but I think the question still stands.
Edit 2: I have either transcribed the problem wrong twice (or more), or the 1st edition copy of the book I have at home is different from the 2nd edition copy online. Either way, the question is flawed. I have asked the corrected question again here.
 A: Your calculations are correct. The thing on top is indeed $-\epsilon$, and the thing at the bottom is
$$\sqrt{\epsilon}(3+\sqrt{9+\epsilon}).$$ 
Note that the infinitesimal $\epsilon$ is positive, since $\sqrt{c-7}$ is not defined when $\epsilon$ is negative.
There is partial cancellation of the $\epsilon$ stuff, since the bottom can be rewritten as $\sqrt{\epsilon}(3+\sqrt{9+\epsilon})$.  After cancelling, on top we get $-\sqrt{\epsilon}$, and at the bottom we get $3+\sqrt{9+\epsilon}$. T 
Remark: The limit of our expression as $c$ approaches $7$ from above is not $-\frac{1}{6}$. If the back of the book says it is, then the back of the book is wrong. Or perhaps you inadvertently typed the expression incorrectly. In fact, the limit is $0$, for the standard part of $\sqrt{\epsilon}$ is $0$, and the standard part of $3+\sqrt{9+\epsilon}$ is $6$.
Let's check numerically. Put $\epsilon=10^{-6}$ (sorry, I have a cheap calculator, it doesn't handle infinitesimals). Calculate our function with $c=7+\epsilon$. The result is roughly $-1.667\times 10^{-4}$, nowhere near $-\frac{1}{6}$.
Note: This post has been rewritten in order to deal with an edit that changed the function.
