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Proof of L'Hôpital's Rule

  1. Mar 15, 2014 #1
    I am somewhat confused by the proof of L'Hôpital's Rule in Pugh's "Real Mathematical Analysis." (See Attachments, Theorem 6). I follow every bit of the proof save one choice and its implication. That is, why choose t based on

    ##\displaystyle |f(t)| + |g(t)| < \frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)}##
    ##\displaystyle |g(t)| < \frac{g(x)}{2}##,

    and how does it follow that

    ##\displaystyle \left |\frac{g(x)f(t) - f(x)g(t)}{g(x)(g(x)-g(t))} \right | < \frac{\epsilon}{2} ##?

    I understand that t need be chosen to be much closer to b than x, but I have yet to be able to convince myself why this meets that goal. Thanks for the help in advance!
     

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    Last edited: Mar 15, 2014
  2. jcsd
  3. Mar 15, 2014 #2

    jbunniii

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    I assume you're OK up to this point:
    $$\left|\frac{g(x)f(t) - f(x)g(t)}{g(x)(g(x)-g(t))} \right|+ \left|\frac{f'(\theta)}{g'(\theta)} - L \right|$$
    Let's work with the numerator of the first term:
    $$\begin{align}
    |g(x)f(t) - f(x)g(t)| &\leq |g(x)f(t)| + |f(x)g(t)| \\
    &\leq |g(x)f(t)| + |f(x)g(t)| + |f(t)f(x)| + |g(t)g(x)| \\
    &= (|f(t)| + |g(t)|)(|f(x)| + |g(x)|) \\
    &< \frac{g(x)^2 \epsilon}{4}\\
    \end{align}$$
    where the last inequality follows because of the first inequality on page 144.
    Using this, we obtain the following bound:
    $$\begin{align}
    \left|\frac{g(x)f(t) - f(x)g(t)}{g(x)(g(x)-g(t))} \right| &\leq
    \left| \frac{g(x)^2 \epsilon}{4g(x)(g(x)-g(t))} \right| \\
    &= \left| \frac{g(x) \epsilon}{4(g(x) - g(t))} \right|\\
    &= \left| \frac{\epsilon}{4(1 - g(t)/g(x))} \right|
    \end{align}$$
    Now by the triangle inequality,
    $$|1 - g(t)/g(x)| \geq 1 - |g(t)/g(x)|$$
    and since ##|g(t)| < |g(x)| / 2##, we have
    $$1 - |g(t)/g(x)| > 1 - 1/2 = 1/2$$
    Thus
    $$\left| \frac{\epsilon}{4(1 - g(t)/g(x))} \right|< \left| \frac{2\epsilon}{4}\right| = \frac{\epsilon}{2}$$
     
    Last edited: Mar 15, 2014
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