
#1
Oct106, 09:10 PM

P: 19

Hello,
I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks. Prove that lim x>2 (x^21)=3 0<f(x)L<epsilon whenever 0<xa<delta 0<(x^21)3<epsilon whenever 0<x(2)<delta = x^24<epsilon = (x2)(x+2)<epsilon = x2x+2<epsilon = If x2<C (C=constant) then x2x+2<Cx+2 = Cx+2<epsilon = x+2<epsilon/C Assume x+2<1 so 1<x+2<1 = 3<x<1 = 5<x2<3 After this step 5<x2<3, I'm inclined to set x2<3, but book states that x2<5. I'm lost at this step, should it be5? Anyways, any explanations would be great. Thanks, Chris 



#2
Oct206, 04:07 AM

Sci Advisor
HW Helper
P: 2,004

This implies that x2<5, simply because x2<3 means that surely x2<5, so that 5<x2<5, or x2<5. You wrote x2<3, which is never true becuase the left side is positive and the right side isn't. Did you mean x2<3? 



#3
Oct206, 09:51 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

Small detail that is important:
In other words, f(x) can be = L. 



#4
Oct206, 09:50 PM

P: 4

Precise definition of a limit
Perhaps this way would work:
Prove that lim x>2 (x^21)=3 let E > 0 be given and: f(x)  L = x^2  4 = (x  2)(x + 2) = x  2x + 2 if 0 < x + 2 < 1 then, 3 < x < 1 x  2 < 5 < 5x + 2 therefore: if 0 < x + 2 < 1 and 5x + 2 < E then by transitivity of <(less than), f(x)  L < E or reworded: if D = minimum(1, E/5), then f(x)  L < E (where E = epsilon; D = delta) 



#5
Oct206, 09:55 PM

P: 4

Sorry this part:
if 0 < x + 2 < 1 then, 3 < x < 1 x  2 < 5 is a "side part" 



#6
Oct306, 12:10 AM

P: 19




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