# Proof of L'Hôpital's Rule

1. Mar 15, 2014

### Someone2841

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. Mar 15, 2014

### jbunniii

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