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!
 A: Everything is fine, apart from the fact that you don't think it is fine.
First, a small piece of advice. Don't multiply unless you have to. At a certain stage you had
$$\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})},$$
which you expanded (but certainly did not simplify) to
$$=\frac{9-(c+2)}{3c+c\sqrt{c+2}-7\sqrt{c+2}-21}.$$
This is correct but looks worse to me. Instead, simplify the top to $7-c$, which cancels with the $c-7$ at the bottom to give $-1$. Or if you really want to, note that $7-c=-\epsilon$, and $c-7=\epsilon$, and cancel the $\epsilon$ (or, to be fancy, divide top and bottom by $\epsilon$). 
You managed, however, to get through the mess to get the right expression
$$-\frac{\epsilon}{3\epsilon + \epsilon\sqrt{9+\epsilon}},$$
but there you seemed to be stuck. 
Note then that the two bottom terms have a common factor of $\epsilon$. "Take it out" to get $\epsilon(3+\sqrt{9+\epsilon})$, and cancel with the $\epsilon$ on top.
However we do these things, we end up with
$$-\frac{1}{3+\sqrt{9+\epsilon}}.$$
Find the standard part of this. The standard part of $9+\epsilon$ is $9$, so the standard part of $\sqrt{9+\epsilon}$ is $3$, so the standard part of the whole thing is $-\dfrac{1}{6}$. 
A: 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.
