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

    $\displaystyle \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:

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

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

    Differentiating wrt $\displaystyle x$:

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

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

    And so on, so that:

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

    and similarly,

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

    So:

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

    $\displaystyle \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 $\displaystyle M \geq N \geq 1 $, then:

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

    $\displaystyle \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 $\displaystyle M \geq N \geq 1 $, then:

    $\displaystyle \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 $\displaystyle M = N $, as $\displaystyle x \to a $, $\displaystyle (x - a)^{M - N} \to 0 $ and so $\displaystyle \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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