
#1
Mar407, 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? 



#2
Mar407, 06:36 PM

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#3
Mar407, 07:01 PM

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. 



#4
Mar407, 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?




#5
Mar407, 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.




#6
Mar407, 08:11 PM

P: 43

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




#7
Mar407, 11:40 PM

P: 372

Define x_n=1/n. This is a sequence converging to 0. But 1/x_n1/x_n=2n which doesn't converge to anything. Therefore 1/x1/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
Mar507, 03:33 AM

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P: 3,353

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. 



#9
Mar507, 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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