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Confirm Precise Definition of a Limit solution 
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#1
Jul308, 03:26 AM

P: 40

As the subject says, I have "proved" a function, but because this type of problem seems to have no set answer, I'd like some opinions on whether I walked through it properly or not.
1. The problem statement, all variables and given/known data lim(x=> 2) (x^21) = 3 2. Relevant equations Precise Definition of a Limit 3. The attempt at a solution Part 1. Assume a value for [tex]\delta[/tex] Since 0 < x + 2<[tex]\delta[/tex], f(x)  3 < [tex]\epsilon[/tex] f(x)  3 = x^2  1  3 = x^2  4 = x + 2x  2< [tex]\epsilon[/tex] Let C = x  2, which leads to Cx + 2 < [tex]\epsilon[/tex] x + 2 = [tex]\delta[/tex] < [tex]\frac{\epsilon}{C}[/tex] Applying the Precise Definition of a Limit, For a [tex]\delta[/tex] value [tex]\frac{\epsilon}{C}[/tex] greater than zero, there exists [tex]\epsilon[/tex] greater than zero such that if 0 < x + 2<[tex]\delta[/tex], then f(x)  3 < [tex]\epsilon[/tex]. f(x)  3 < [tex]\epsilon[/tex] f(x)  3 = x^2  1  3 = x^2  4 = x + 2x  2< [tex]\epsilon[/tex] Reapply C and [tex]\delta[/tex] to get C x [tex]\frac{\epsilon}{C}[/tex] < [tex]\epsilon[/tex] Hence, by the Precise Definition of a limit, said limit does exist. Much thanks in advance. Comments and criticisms are always welcome. 


#2
Jul308, 03:55 AM

P: 152

It looks like you're doing this backwards. Your proof should start like "Given [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]0<x+2<\delta[/itex], then [itex]f(x)3<\epsilon[/itex]..." Now show that this is true using the $\delta$ that you found.



#3
Jul308, 04:55 AM

Math
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Thanks
PF Gold
P: 39,491

what you need is a BOUND on x2. If C is close to 2, say between 1 and 3, how large or how small can x2 be? 


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