1
$\begingroup$

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!

2 Answers 2

2

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 $-\dfrac{1}{6}$.

  • 1
    Please don't misunderstand me; I'm a programmer, I'm well aware of how important it is to keep things simple. :) If I'm doing something in an obtuse way when things are not complex, how much worse will it be when things are complex?2012-01-31
1

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.

  • 0
    Thanks! It seems obvious in retrospect!2012-01-31