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Prove limit x approaches 0 of a rational function = ratio of derivatives

  1. Oct 23, 2014 #1
    1. The problem statement, all variables and given/known dat
    If f and g are differentiable functions with f(O) = g(0) = 0 and g'(O) not equal 0, show that

    lim f(x) = f'(0)
    x->0 g(x) g'(0)

    3. The attempt at a solution
    I know that lim as x→a f(a) = f(a) if function is continuous. since its differentiable it's continuous. so lim x→0 f(x) = f(0). and lim x→0 g(x) = g(0) but you can't have a 0/0.
    I have so idea how to get to the derivative part.
     
  2. jcsd
  3. Oct 23, 2014 #2

    PeroK

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    This is essentially a special case of L'Hospital's rule. Do you know the Mean Value Theorem?
     
  4. Oct 23, 2014 #3
    Yes. but how does mvt apply? we didn't learn lhopitals rule yet.
     
  5. Oct 23, 2014 #4

    PeroK

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    Actually, you don't even need the MVT. Just think about the definition of a derivative.
     
  6. Oct 23, 2014 #5

    RUber

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    This is one of those cases where adding zero back into the expression makes it more clear:
    ##\lim_{a\to 0} f(a) = \lim_{a\to 0} f(0+a)-f(0) ## ...
     
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