# An Error in Spivak's Calculus?

1. Aug 10, 2011

### middleCmusic

An Error in Spivak's Calculus? [problem resolved]

EDIT: Thanks for your help, guys. I see where I went wrong.

At one point in the discussion of limits, he says:

If |x-a| < 1, then |x| < |a| + 1.

As far as I can tell, this is true if x and a are both positive, but not otherwise.

Am I missing something here?

(And no, there isn't really any other information regarding the x-value or the value of a.)

Last edited: Aug 11, 2011
2. Aug 10, 2011

### disregardthat

It's true. |x|-|a| <= |x-a| < 1, so |x| < |a| + 1

3. Aug 10, 2011

### gb7nash

This is a result of the triangle inequality. We know for any real numbers a,b that:

|a-b| >= |a| - |b| (do you know how to prove this using triangle inequality?)

Now looking back at the problem, Since |x-a| < 1, this implies that |x| - |a| <= |x-a| < 1, so...

4. Aug 10, 2011

### middleCmusic

Wow, thanks for clearing that up. I'm not sure what I was thinking before, but it makes sense now.

5. Aug 11, 2011

### pessimist

|x - a| < 1
-1 + a < x < 1 + a
|x| < |1 + a| <= 1 + |a|

6. Aug 11, 2011

### middleCmusic

Thanks, your explanation is terrific.

7. Aug 11, 2011

### Yuqing

This is quite a common trick in epsilon delta proofs.

|x| = |x + a - a| <= |x - a| + |a| < |a| + 1

this version requires only the standard triangle inequality without the need to break the absolute value up.

8. Aug 11, 2011

### pessimist

I just realised there is an error in my proof :) sorry about that.

|x| = |x + a - a| <= |x - a| + |a| < 1 + |a|

This one posted above is abolutely correct.

9. Aug 11, 2011

### pessimist

I only meant to do it so that a better understanding was possible but I erred.

10. Aug 18, 2011

### Aldaron

Isn't there an error (or some sloppiness, at least) in the subsequent paragraph though: shouldn't

|x-a| < ε/(2|a|+1)​

be

|x-a| ≤ ε/(2|a|+1)​

11. Aug 18, 2011

### disregardthat

In what context? What subsequent paragraph are you talking about, what does it say?

12. Aug 19, 2011

### Aldaron

On the same page (93) as the rest of what's being discussed here (text line 18).

13. Aug 19, 2011

### Yuqing

I think you're right in that less than or equal to is sufficient. But when discussing epsilon - delta limits, it is simply easier to use strict inequalities without changing the end result too much.

14. Aug 19, 2011

### disregardthat

No, it's <. We require |x-a| < ε/(2|a|+1). You should copy the text, not everyone has the textbook.

15. Aug 19, 2011

### Aldaron

Yes, that's exactly what's going on. Thanks.

16. Aug 19, 2011

### Aldaron

The question is about the textbook (as your answer demonstrates).