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Finding the limit of the equation

by shwanky
Tags: equation, limit
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shwanky
#1
Mar4-07, 06:09 PM
P: 43
1. The problem statement, all variables and given/known data
Evaluate the limit if it exists

2. Relevant equations

[tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}[/tex]


3. The attempt at a solution

1) [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}[/tex]

2a) [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{x}[/tex]

2b) [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{-x}[/tex]

3a) [tex]\lim_{x\to 0} 0 = 0[/tex]

3b) [tex]\lim_{x\to 0} \frac{1}{2x}[/tex]

4)[tex]\lim_{x\to 0} \frac{1}{2}x[/tex]

5)[tex] (\frac{1}{2})0 = 0[/tex]

Did I do this correctly?
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Dick
#2
Mar4-07, 06:36 PM
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Quote Quote by shwanky View Post
Did I do this correctly?
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.
ZioX
#3
Mar4-07, 07:01 PM
P: 371
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.

shwanky
#4
Mar4-07, 07:07 PM
P: 43
Finding the limit of the equation

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?
Dick
#5
Mar4-07, 07:10 PM
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Yes. The limits from the two sides are different. So there is no common limit. So the limit doesn't exist.
shwanky
#6
Mar4-07, 08:11 PM
P: 43
OMG! I can't believe it was that simple... :(
ZioX
#7
Mar4-07, 11:40 PM
P: 371
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.
Gib Z
#8
Mar5-07, 03:33 AM
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For the limit to exist, [tex]\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|}[/tex]

However we can see that [tex] \lim_{x\to 0^{+}} \frac{1}{x}-\frac{1}{|x|}=0[/tex] but [tex]\lim_{x\to 0^{-}} \frac{1}{x}-\frac{1}{|x|} = \lim_{x\rightarrow 0} \frac{-2}{x}[/tex], which are obviously not the same.
HallsofIvy
#9
Mar5-07, 05:49 AM
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[tex]\frac{1}{2x}\ne \frac{1}{2} x[/tex]!!!

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


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