Expert: Paul Klarreich - 1/30/2012

Question
Hi Sir,

  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
lim[x->a] x^1/3 = a^1/3


Let y = x^1/3,  b = a^1/3

Show that given  e [epsilon], we can find d [delta] such that

| x^1/3 - a^1/3 | < e  whenever  | x - a | < d


| x - a | < d   means:


| y^3 - b^3 | < d

| (y - b)(y^2 + by + b^2) | < d

Now

| x - a | < d   means:

- d < x - a < d

a - d < x < a + d

Assume we will always take  d < a/2.  Then

x > a/2,

y > (a/2)^1/3 = b/2^1/3


y^2 + by + b^2 > b^2(1/2^2/3 + 1/2^1/3 + 1) = b^2(M)


| (y - b)(y^2 + by + b^2) | < d


| (y - b) | < d/(y^2 + by + b^2) < d/[b^2(M)]

Now take  d < e b^2(M) and

| (y - b) | < e