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Math Help - Proving L'Hopital's (Bernoulli's) Rule

  1. #1
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    Question Proving L'Hopital's (Bernoulli's) Rule

    I wish to prove the following.

    "Let  f(x), g(x) be represented by Taylor series at x = a , where f(a) = g(a) = 0 , and let the first non-vanishing derivatives at  a be f^{(M)} (a) and g^{(N)}(a) , where  M \geq N \geq 1 . Then

     \displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f^{(M)}(a)}{g^{(N)}(a)} "

    So here's how I've proceeded so far:

    f(x) = A_0 + A_1 (x - a) + A_2 (x - a)^2 + \ldots

    If we let  x = a , then  A_0 = f(a)

    Differentiating wrt x:

    f'(x) = 1 \cdot A_1 + 2 \cdot A_2 (x - a)^1 + \ldots

    Again, letting  x = a , then  A_1 = \frac{1}{1!}f'(a)

    And so on, so that:

    f(x) = f(a) + \frac{1}{1!} f^{(1)}(a) (x - a) + \frac{1}{2!} f^{(2)} (a) (x - a)^2 + \ldots

    and similarly,

     g(x) = g(a) + \frac{1}{1!} g^{(1)} (a) (x - a) + \frac{1}{2!} g^{(2)} (a) (x - a)^2 + \ldots

    So:

     \frac{f(x)}{g(x)} = \frac{ f(a) + \frac{1}{1!} f^{(1)} (a) ( x - a ) + \frac{1}{2!} f^{(2)} (a) (x - a)^2 + \ldots }{g(a) + \frac{1}{1!} g^{(1)} (a) (x - a) + \frac{1}{2!} g^{(2)} (a) (x - a)^2 + \ldots}

    Recalling that  f^{(M)} (a) and g^{(N)} (a) are the first non-vanishing derivatives at  a . Then:

     \frac{f(x)}{g(x)} = \frac{\frac{1}{M!} f^{(M)} (a) (x - a)^M + \textrm{higher powers}}{\frac{1}{N!} g^{(N)} (a) (x - a)^N + \textrm{higher powers}}

    Recalling that  M \geq N \geq 1 , then:

     \frac{f(x)}{g(x)} = \frac{\frac{1}{M!} f^{(M)} (a) (x - a)^{M-N} + \textrm{higher powers}}{\frac{1}{N!} g^{(N)} (a) + \textrm{higher powers}}

    But unless  M = N , as x \to a ,  (x - a)^{M - N} \to 0 and so  \frac{f(x)}{g(x)} \to 0 , which clearly is not the desired result.

    Where am I going wrong?
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  2. #2
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    Quote Originally Posted by Harry1W View Post
    I wish to prove the following.

    "Let  f(x), g(x) be represented by Taylor series at x = a , where f(a) = g(a) = 0 , and let the first non-vanishing derivatives at  a be f^{(M)} (a) and g^{(N)}(a) , where  M \geq N \geq 1 . Then

     \displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f^{(M)}(a)}{g^{(N)}(a)} "

    So here's how I've proceeded so far:

    ...
    [SNIP]
    ...

    Recalling that  M \geq N \geq 1 , then:

     \frac{f(x)}{g(x)} = \frac{\frac{1}{M!} f^{(M)} (a) (x - a)^{M-N} + \textrm{higher powers}}{\frac{1}{N!} g^{(N)} (a) + \textrm{higher powers}}

    But unless  M = N , as x \to a ,  (x - a)^{M - N} \to 0 and so  \frac{f(x)}{g(x)} \to 0 , which clearly is not the desired result.

    Where am I going wrong?
    You're not going wrong. The "desired result" is not true unless M=N. As your analysis correctly shows, the limit will be zero if M>N.
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