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Limit problem

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data

    If ##\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1##.
    then f'(1) = ...

    A. -1
    B. -1/3
    C. 1/3
    D. 1
    E. 2

    2. Relevant equations

    ##f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}##


    3. The attempt at a solution

    I really have no idea about this problem.
    Please help me.
     
  2. jcsd
  3. Apr 12, 2016 #2

    PeroK

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    First, rewrite the definition of the derivative limit. There are always two alternative ways o wrtite a limit. And, I'm not talking about epsilon-delta.

    Second, a hint: you need to learn to spot a composite function when you see one!
     
  4. Apr 12, 2016 #3
    Do you mean the L' Hospital rule?
    By using L'Hospital rule, I get
    ##\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} = -1##
    ##\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1##
    ##\lim_{x->a}\frac{3x^2 f'(x^3))}{x}=-1##
    ##3a^2 f'(a^3) = -a##
    ##f'(a^3) = \frac{-a}{3a^2}##
    ##f'(a^3) = -\frac{1}{3a}##
    ##f'(1^3) = -\frac{1}{3}##

    Is it correct ?
     
  5. Apr 12, 2016 #4

    PeroK

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    An interesting solution! That's probably valid, but it's not what I meant. Note that:

    ##f'(a) \equiv \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a} \equiv \lim_{h\rightarrow 0} \frac{f(a + h)-f(a)}{h}##

    Those are the two equivalent ways of writing the definition of a derivative.

    And, if you let ##g(x) = f(x^3)##, then what you were given was effectively ##\forall a, \ g'(a) = -1##. The rest follows from the chain rule.
     
  6. Apr 12, 2016 #5
    g'(a) = -1
    3a^2 f'(a^3) = -1
    We let a = 1
    f'(1)= -1/3
    Thanks for your help
     
  7. Apr 12, 2016 #6

    mfb

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    As the problem is solved, here is a different approach: Define ##y=x^3##, ##b=a^3##.
    Then
    $$-1 = \lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} \\
    = \lim_{y\rightarrow b} \frac{f(y)-f(b)}{\sqrt[3]{y}-\sqrt[3]{b}} \cdot \frac{\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2}{\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2} \\
    = \lim_{y\rightarrow b} \frac{f(y)-f(b)}{y-b} \cdot \left(\sqrt[3]{y}^2+\sqrt[3]{y}\sqrt[3]{b}+\sqrt[3]{b}^2\right)$$
    The fraction goes to f'(b), the second part converges to 3. Plug in b=1, done.
     
  8. Apr 12, 2016 #7

    SammyS

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    Similarly:

    ##\displaystyle \lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x-a} \ ##

    ##\displaystyle =\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x^3-a^3}\frac{x^3-a^3}{x-a} \ ##

    ##\displaystyle =\lim_{x\rightarrow a} \frac{f(x^3)-f(a^3)}{x^3-a^3}\cdot \lim_{x\rightarrow a}\frac{(x-a)(x^2+ax+a^2)}{x-a} \ ##

    ##\displaystyle =f'(a)\cdot \lim_{x\rightarrow a}(x^2+ax+a^2)\ ##​
     
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