How to prove?
- Thread starter hoang1881
- Start date
sorry for my unclearness. I dont know how to explain it in English. I know the epsilon-delta definition (I didn't how it is actually called before your comment). But I just do not know how to use it to prove this function (maybe I have to set an epsilon or a max or min number but im not sure). I can only prove some really basic functions.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Preliminaries: Suppose that \(\varepsilon>0,~(x^3-27)=(x-3)(x^2+3x+9)~ \&~|x-3|<1\) Then
\(-1<x-3<1\\2<x<4\\6<3x<12\\4<x^2<16\\10<x^2+3x<28\\19<x^2+3x+9<37\)
Now we build the limit. Suppose that \(\delta= \min\left\{1,\dfrac{\varepsilon}{37}\right\}\)
So if \(|x-3|<1\) then \(|x^2+3x+9|<37|\).
If \(|x-3|<\delta\) then it follows that \(\left|x^3-27\right|=\left|x-3\right|\cdot\left|x^2+3x+9\right|<\varepsilon\)
\(-1<x-3<1\\2<x<4\\6<3x<12\\4<x^2<16\\10<x^2+3x<28\\19<x^2+3x+9<37\)
Now we build the limit. Suppose that \(\delta= \min\left\{1,\dfrac{\varepsilon}{37}\right\}\)
So if \(|x-3|<1\) then \(|x^2+3x+9|<37|\).
If \(|x-3|<\delta\) then it follows that \(\left|x^3-27\right|=\left|x-3\right|\cdot\left|x^2+3x+9\right|<\varepsilon\)