# Finding the limit of the equation

1. Mar 4, 2007

### shwanky

1. The problem statement, all variables and given/known data
Evaluate the limit if it exists

2. Relevant equations

$$\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}$$

3. The attempt at a solution

1) $$\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}$$

2a) $$\lim_{x\to 0} \frac{1}{x}-\frac{1}{x}$$

2b) $$\lim_{x\to 0} \frac{1}{x}-\frac{1}{-x}$$

3a) $$\lim_{x\to 0} 0 = 0$$

3b) $$\lim_{x\to 0} \frac{1}{2x}$$

4)$$\lim_{x\to 0} \frac{1}{2}x$$

5)$$(\frac{1}{2})0 = 0$$

Did I do this correctly?

Last edited: Mar 4, 2007
2. Mar 4, 2007

### Dick

The limit is quite different if you approach zero from the negative direction or the positive direction. Split it into these two cases as I think you have tried to do, but 1a) 1b) 2a) 2b) etc is not the clearest way to express this. But finally 1/(2*x) is not equal to (1/2)*x. That's BAD.

3. Mar 4, 2007

### ZioX

Well, you didn't do it right as it doesn't exist.

1/x goes to -\infty as x->0 from the left. but -1/|x| is also going to -\infnty.

Use the sequence criterion of limits for a more formal proof.

Also note 1/x + 1/x is not 1/2x.

Last edited: Mar 4, 2007
4. Mar 4, 2007

### shwanky

Thanks, I kind of figured it was wrong. I'm still a little shakey on limits. If it's a problem that I can factor and substitute, I'm okay, but my professor didn't clearly explain this in class. I can usually see that the limit doesn't exist, but I don't know how to state it. Should I just take the + and - limits and show that they don't approach a common point on an open interval as ZioX did?

5. Mar 4, 2007

### Dick

Yes. The limits from the two sides are different. So there is no common limit. So the limit doesn't exist.

6. Mar 4, 2007

### shwanky

OMG! I can't believe it was that simple... :(

7. Mar 4, 2007

### ZioX

Define x_n=-1/n. This is a sequence converging to 0. But 1/x_n-1/|x_n|=-2n which doesn't converge to anything. Therefore 1/x-1/|x| has no limit as x tends to zero.

This is the sequencial criterion for limits. A function f converges to L as x converges to a iff for every sequence x_n converging to a has the property that f(x_n) converges to L.

8. Mar 5, 2007

### Gib Z

For the limit to exist, $$\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|} =\lim_{x\to 0^{+}} \frac{1}{x}-\frac{1}{|x|} = \lim_{x\to 0^{-}} \frac{1}{x}-\frac{1}{|x|}$$

However we can see that $$\lim_{x\to 0^{+}} \frac{1}{x}-\frac{1}{|x|}=0$$ but $$\lim_{x\to 0^{-}} \frac{1}{x}-\frac{1}{|x|} = \lim_{x\rightarrow 0} \frac{-2}{x}$$, which are obviously not the same.

9. Mar 5, 2007

### HallsofIvy

Staff Emeritus
$$\frac{1}{2x}\ne \frac{1}{2} x$$!!!

And, technically, you should say $lim_{x\rightarrow 0^+}$ and $lim_{x\rightarrow 0^-}$ but that isn't as bad as the howler above!