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I have to prove x^3 is differentiable at x=4

  1. Dec 11, 2007 #1
    I have to prove [tex]x^3[/tex] is differentiable at [tex]x=4[/tex] using the definition of what it means for something to be differentiable.

    So I was wondering if I just have to show that [tex]f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4}[/tex] exists, where [tex]f(x) = x^3[/tex].


    [tex]f'(4) = \lim_{x \to 4} \frac {x^3 - 64}{x - 4}[/tex]

    [tex]= \lim_{x \to 4} \frac{(x-4)(x^2 + 4x + 16)}{x-4}[/tex]

    [tex] = \lim_{x \to 4} x^2 + 4x + 16[/tex].

    So the limit (derivative) is 48...does this prove it's dif'able at [tex]x=4[/tex]?
  2. jcsd
  3. Dec 11, 2007 #2


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    The proof is acceptable at an elementary calculus level. If you are taking advanced calculus, you would need a little more formalism, but the idea is the same.
  4. Dec 11, 2007 #3
    Ha, then I bet for a Real Analysis course this proof would be unacceptable :tongue:
  5. Dec 12, 2007 #4
    actually all you need to do is add a few lines of text justifying each step. Use facts like continuity preserves convergence and the functional limit at a point x does not even require the function to be defined at the point, just near it.

    That aside the proofs for adcalc/analysis and calculus are identical
  6. Dec 12, 2007 #5
    With a few more epsilons and deltas in the mix ;)
  7. Dec 12, 2007 #6
    once you prove that functional limit definition is equivalent to the sequence limit definition (this is analogous to continuity and sequential continuity the difference being you don't consider the point which you are approaching, in fact f doesn't have to be defined at that point for the limit to exist) you see that because of continuity

    the limit of a polynomial at x is the polynomial evaluated at x.

    so that any epsilon-delta argument is unnecessary.
  8. Dec 13, 2007 #7


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    Because "differentiable at x= a" means precisely that that limit exists!
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