Expert: Frederick Koh - 1/30/2012

Question
Hello Mr Frederick Koh,

  I'm from NUS too & currently in Year 1. I would like to seek your advice to solve this problem:

  Prove, using epsilon-delta definition, (x)^1/3 = a^1/3, as x tends to a>0.

  Here's my working(s):

mod(x-a) < delta => mod(x^1/3  - a^1/3) < epsilon

mod (x-a) = mod {(x^1/3 - a^1/3)(x^[2/3] + (ax)^[1/3] + a^[2/3])}

         < mod(x^1/3 - a^1/3)[mod (x^[2/3]) + mod((ax)^[1/3]) + mod(a^[2/3]) }


Alfredo

Answer
The essence of epsilon-delta is to obtain a relationship between epsilon and delta themselves.
If you can achieve this, then the limit proof falls nicely into place.

From your workings,

mod (x-a) = mod {(x^1/3 - a^1/3)(x^[2/3] + (ax)^[1/3] + a^[2/3])}
         mod (x^1/3 - a^1/3)* mod (x^[2/3] + (ax)^[1/3] + a^[2/3])-------------(1)

Letting mod (x^[2/3] + (ax)^[1/3] + a^[2/3])=k,    where k clearly is >0
(1) becomes     mod (x-a) = mod (x^1/3 - a^1/3)* k
         which is lesser than epsilon * k

Hence, we can establish the relationship that  epsilon * k=delta, which means
(x)^1/3 = a^1/3, as x tends to a>0.  (shown)


Note: your last line of workings
         mod(x^1/3 - a^1/3)[mod (x^[2/3]) + mod((ax)^[1/3]) + mod(a^[2/3]) }
is not correct; you cannot simply tear apart x^[2/3] + (ax)^[1/3] + a^[2/3] and assign the modulus operator to the 3 resulting separate terms.
       

Hope this helps. Peace.