Manipulating Limits

Date: 10/13/2003 at 21:45:29
From: Dijana 
Subject: calculus

I have a three-part question:

(a) If g(x)=x^(2/3), show that g'(0) does not exist.
(b) If a is not equal to 0, find g'(a).
(c) Show that y=x^(2/3) has a vertical tangent line at (0,0).

I used the formula f'(a) = lim x=>a where f(x)-f(a)/(x-a), but I don't 
understand what to do with the question.

Date: 10/14/2003 at 07:10:21
From: Doctor Luis
Subject: Re: calculus

Hi Dijana,

If we look carefully at

                 x^(2/3) - a^(2/3)
 f'(a) = lim   --------------------
         x->a          x - a


we see that we can't just evaluate it at x=a because we get division
by zero. So, we must find a way to make the top cancel out the bottom.

First of all, the numerator looks like a difference of squares.
That seems to be a big hint.

  x^(2/3) - a^(2/3) = (x^(1/3) - a^(1/3)) * (x^(1/3) + a^(1/3))

So, we have to get cube roots out of the denominator somehow.
We can do this by writing x as (x^(1/3))^3 and 'a' as (a^(1/3))^3.
This means that x-a can be written as a difference of cubes:

  x - a = (x^(1/3))^3 - (a^(1/3))^3

And what do we know about differences of cubes? Well, there's
the identity

  m^3 - n^3 = (m - n) * (m^2 + m*n + n^2)

which means we can rewrite x-a as

  x - a = (x^(1/3) - a^(1/3)) * (x^(2/3) + (ax)^(1/3) + a^(2/3))

Do you see that first factor? It's also a factor in the numerator
for our limit!

To summarize,

                 x^(2/3) - a^(2/3)
 f'(a) = lim   --------------------
         x->a          x - a

               (x^(1/3) - a^(1/3)) * (x^(1/3) + a^(1/3))
       = lim   ------------------------------------------------------
         x->a  (x^(1/3) - a^(1/3)) * (x^(2/3) + (ax)^(1/3) + a^(2/3))

                    (x^(1/3) + a^(1/3))
       = lim   --------------------------------
         x->a  (x^(2/3) + (ax)^(1/3) + a^(2/3))

Taking the limit should now be straightforward. 

I hope this helped!  Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/