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Limit Of A Function From Definition 
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#1
Oct2009, 06:56 AM

P: 82

I have to prove from the definiton of the limit of a function that the limit of f(x) as x tends to 2 equlas 3, given that;
f(x)=(2x1) I know that I have to find an epsilon such that f(x)l [tex]\leq[/tex] [tex]\epsilon[/tex] and delta such that 0 [tex]\leq[/tex] xa [tex]\leq[/tex] [tex]\delta[/tex] Nowing putting in the conditions for this f(x); 2x4 [tex]\leq[/tex] [tex]\epsilon[/tex] and 0 [tex]\leq[/tex] x2 [tex]\leq[/tex] [tex]\delta[/tex] But I don't where to go from here. Any help would be great! 


#2
Oct2009, 07:06 AM

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Well, you have:
2x4=2x2 Can this be made arbitrarily small (i.e, below some fixed "epsilon"), as long as we pick a small enough "delta"? In particular, IF we want [itex]2x2\leq\epsilon[/itex], what is the largest value we can pick for "delta" in order to ENSURE the validity of the above inequality? 


#3
Oct2009, 07:54 AM

P: 49

Often, we need to find [tex]\delta = \delta(\epsilon)[/tex] as a function of the [tex]\epsilon[/tex] we were given.
That is, write [tex]f(x)l[/tex] as a function [tex]xa[/tex] and use your delta to show that it is less than epsilon 


#4
Oct2009, 12:43 PM

P: 82

Limit Of A Function From Definition
So I know that f(x)l=2(x2)=2xa
Would it then be right to say that since xa=x2 [tex]\leq[/tex] [tex]\delta[/tex] then f(x)l=2x2 [tex]\leq[/tex] 2[tex]\delta[/tex] hence taking 2[tex]\delta[/tex] [tex]\leq[/tex] [tex]\frac{\epsilon}{2}[/tex] would satisfy the problem?? 


#5
Oct2009, 01:04 PM

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(you meant, I think delta<=epsilon/2, I think..) 


#6
Oct2009, 01:25 PM

P: 82

Yeah that is what I meant! Woops. So is the following proof correct then;
Definition of the Limit of a Function at a Point. Suppose that f(x) is defined on (aR,a)U(a,a+R), for some R [tex]\succ[/tex] 0. Then f(x) tends to l as x tends to a if, given any [tex]\epsilon[/tex] in R[tex]^{+}[/tex] , there exists [tex]\delta[/tex] in R[tex]^{+}[/tex] such that, f(x)l [tex]\prec[/tex] [tex]\epsilon[/tex] whenever 0 [tex]\prec[/tex] xa [tex]\prec[/tex] [tex]\delta[/tex] Now in this case f(x)=2x1, l=3 and a=2 Hence 2x4 [tex]\prec[/tex] [tex]\epsilon[/tex] Now 2x2 [tex]\prec[/tex] [tex]\epsilon[/tex] = xa[tex]\prec[/tex] [tex]\delta[/tex] Hence taking [tex]\delta[/tex] [tex]\leq[/tex] [tex]\frac{\epsilon}{2}[/tex] Then f(x)l [tex]\prec[/tex] [tex]\epsilon[/tex] whenever 0 [tex]\prec[/tex] xa [tex]\prec[/tex] [tex]\delta[/tex] As required 


#7
Oct2009, 01:41 PM

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Slightly scruffylooking proof, but okay nonetheless!



#8
Oct2009, 01:43 PM

P: 82

Thanks for the help!



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