# Finding the limit of the equation

by shwanky
Tags: equation, limit
 P: 43 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?
HW Helper
Thanks
P: 24,975
 Quote by shwanky 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.
 P: 372 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.
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?
 Sci Advisor HW Helper Thanks P: 24,975 Yes. The limits from the two sides are different. So there is no common limit. So the limit doesn't exist.
 P: 43 OMG! I can't believe it was that simple... :(
 P: 372 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.
 HW Helper P: 3,353 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,705 $$\frac{1}{2x}\ne \frac{1}{2} x$$!!! And, technically, you should say $lim_{x\rightarrow 0^+}$ and $lim_{x\rightarrow 0^-}$ but that isn't as bad as the howler above!

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