Problem: Show that limx→4x3=64.
My steps:
|f(x)−L|<ϵ so |x3−64|<ϵ
|x3|<64+ϵ→64−ϵ<x3<64+ϵ
$\sqrt[3]{64-\epsilon}
Let ϵ>0. Pick δ so that δ<3√64+ϵ−4.
Suppose 0<|x−4|<δ. Then |x−4|<3√64+ϵ−4.
Thus, |x|<3√64+ϵ and cube to get |x3|<64+ϵ.
Then |x3−64|<ϵ which completes the proof.
But this is what the answer said:
Let ϵ>0. Pick δ so that δ<1 and δ<ϵ61.
Suppose 0<|x−4|<δ.
Then $4−\delta
x3<δ3+12∗δ2+48∗δ+64<δ+12δ+48δ+64
=64+ϵ.
The other inequality is similar.
So my questions are:
1. How can I get δ<ϵ61?
2. Why x3<δ3+12∗δ2+48∗δ+64<δ+12δ+48δ+64=64+ϵ?
3. When we need to pick δ=min(1,ϵ61) in this case?
4. Does my proof has problems ?
I am just a newbie to calculus... Thanks in advance.
Answer
Well, one is not supposed to use inverse functions like cube roots in expression for δ otherwise this leads to a circular argument.
The typical approach is to analyze your goal namely |x3−64|<ϵ
Now we have the obvious inequality |x3−64|=|x−4||x2+4x+16|<61|x−4|
Such problems do not have a unique answer and most importantly they are not supposed to be solved via algebraic manipulation of inequalities.
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