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Help with proving limits using EpsilonDelta definition 
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#1
Oct304, 07:25 AM

P: 3

I am having trouble proving the limits of quadratic functions such as the following. (I used "E" to represent epsilon and "d" for delta)
lim [(x^2)+1] x>1 I found the limit, L, to equal 2 and have proceeded through the following steps: f(x)  L < E  [(x^2)+1]  1 < E [(x^2)1] < E x+1x1 < E While I also know that 0 < x1 < d. My question is how do you find the numerical relationship between x+1x1 and x1 so that I may find d in terms of E? (I was thinking of finding the bounded open interval in which x+1x1 = [(x^2)1] and substituting the greastest figure of the interval, which would be greater than x+1, in place of x+1 so that I would have zx1< E where z is an identified numerical value. However, in problems like the one above, x+1x1 = [(x^2)1] within a seemingly unbounded interval.) please help. your time and assistance is very much appreciated. 


#2
Oct304, 08:02 AM

P: 696

Use the triangle inequality perhaps, x + 1 = x  1 + 2 <= x  1 + 2 < d + 2.



#3
Oct304, 10:23 AM

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What you have done, is to make a GUESS at the limit value!! (As it happens, you've made a correct guess, but that is irrelevant; it's still a guess). What you are to do now, is: 1. Does my guess (2) satisfy the properties that a limit must have? 2. Let [tex]x1<\delta[/tex] 3. Then, [tex]x^{2}+12=x^{2}1=x1x+1[/tex] 4. Now, by assumption, [tex]x1<\delta[/tex] Let us make a further assumption, that [tex]\delta<1[/tex] Then, [tex]x+1<2 (\delta<1)[/tex] And: [tex]x^{2}+12<2\delta, (\delta<1)[/tex] 5. Let [tex]\epsilon>0[/tex] If we are to have [tex]x^{2}+12<\epsilon[/tex] for all x satisfying [tex]x1<\delta[/tex] it is sufficient if the following inequalities are simultaneously satisfied: [tex]2\delta<\epsilon[/tex] [tex]\delta<1[/tex] Hence, setting [tex]\delta=min(\frac{\epsilon}{2},1)[/tex] suffices. That is, we were able to show that our guess (2) satisfy the properties a limit must have. 


#4
Oct404, 04:43 PM

P: 3

Help with proving limits using EpsilonDelta definition
By 'finding' 2 through the substitution of 1 in f(x)=[(x^2)+1] I meant that it was a possible limit whose validity must be varified through the Ed definition. Thank you for correcting me however, to avoid my own future confusion.
Thanks for the helpful guidance and clarification. 


#5
Oct404, 04:47 PM

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I was perhaps a bit too snappish on that point, don't bite me back, though..



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