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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!

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2 Answers 2

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

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Thanks again for your help! I am (obviously, I suppose) trying to teach myself this material and I can use all the aesthetics advice I can get. –  Daniel Lyons Jan 31 '12 at 22:55
    
@Daniel Lyons: It is not a question of "mere" aesthetics, it is practical. Pretty expressions are easier to work with. –  André Nicolas Jan 31 '12 at 23:08
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? –  Daniel Lyons Jan 31 '12 at 23:47

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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Thanks! It seems obvious in retrospect! –  Daniel Lyons Jan 31 '12 at 22:20

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