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:


To get the answer in the back of the book, $-\frac{1}{6}$, I think one must somehow convert this into:


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.

  • $\begingroup$ What is the meaning of $st(c)=7$? $\endgroup$ – emiliocba Jan 31 '12 at 20:50
  • $\begingroup$ It means that $c \in \mathbb{R}^*$ that is infinitely close to 7, that is, the standard part of c is 7. Combining that fact with $c\neq 7$, one concludes that c is 7, plus some infinitesimal value. $\endgroup$ – Daniel Lyons Jan 31 '12 at 20:53

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.

  • $\begingroup$ Problem is, the answer at the back of the book is $-\frac{1}{6}$, not 0, which is what it would be if it were $\frac{\epsilon}{p}$ where $p \in \mathbb{R}, p \neq 0$. $\endgroup$ – Daniel Lyons Jan 31 '12 at 21:15
  • $\begingroup$ @Daniel Lyons: I had already added a remark at the end about the back of the book! We end up with a $\sqrt{|\epsilon|}$ on top, not $\epsilon$, but that's enough to make the limit $0$. $\endgroup$ – André Nicolas Jan 31 '12 at 21:28
  • $\begingroup$ Yes, I transcribed the problem wrong. I just fixed that. However, if you take the limit as c approaches 7 on your calculator you'll get -1/6. Sorry for the confusion! I will try to transcribe more carefully in the future. $\endgroup$ – Daniel Lyons Jan 31 '12 at 21:32
  • $\begingroup$ I have apparently fixed the error by transcribing it wrong AGAIN! $\endgroup$ – Daniel Lyons Jan 31 '12 at 21:45
  • $\begingroup$ I have asked the question again with the corrected problem here. If you are still willing to help me with it, I'd appreciate it a lot! Thanks! $\endgroup$ – Daniel Lyons Jan 31 '12 at 22:09

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