limit proof

Hi,

if you want to prove that limf(x) =4 as x goes to 2 where

f(x)= x^2 if x=/=2 an 1 if x=2 Then

let e>0 be given an we need to find a d>0 such that whenever

0<|x-2|<d we have |f(x)-4|<e. So we solve

|f(x)-4|<e with f(x) = x^2 to give

4-e<x^2<e+4

at which point you take the square root which is assuming e<4.

so you get to

rt(4-e) < x < rt(4+e)

and then we just take d to be the minimum of min(2-rt(4-e) , rt(4+e) -2)

but earlier we assumed that e<4, how does this prove the statement if the definition requires that you can choose any e>0? Does The delta i found work for larger values of e as well? thanks

The formatting is a bit hard to read, but all the defn of a limit requires is that for epsilon > 0, there exists a delta. The bound doesn't have to be tight, so for your example, for all epsilon > 1, you could assume there was a constant ball of size "c" (delta) about 2 and that would be ok.

So is what you are saying that if we chose say e=6 then we would not use my formula for delta with e=6. We would use say e=3 to find the delta with the equation for delta as that would be defined, thanks

So is what you are saying that if we chose say e=6 then we would not use my formula for delta with e=6. We would use say e=3 to find the delta with the equation for delta as that would be defined, thanks