Clarification of L'Hopital Proof Pugh

Question

I am self-studying Real Analysis right now via Pugh's Real Mathematical Analysis but am having trouble understanding a step of the author's proof of L'Hopital's rule.

The theorem is stated as:

If $f$ and $g$ are differentiable functions defined on an intveral $(a,b)$, both of which tend to $0$ at $b$, ad if the ratio of their derivatives $f'(x)/g'(x)$ tends to a finite limit $L$ at $b$ then $f(x)/g(x)$ also tends to $L$ at $b$, where $g(x),g'(x) \neq 0.$

His proof reads as follows:

Given $\epsilon > 0$ we must find a $\delta > 0$ such that if $|x-b| < \delta$ then $|f(x)/g(x) - L|< \epsilon.$ Since $f'(x)/g'(x)$ tends to $L$ as $x$ tends to $b$ there does exist a $\delta > 0$ such that if $x \in (b-\delta, b)$ then $$\left\vert \frac{f'(x)}{g'(x)}-L \right\vert < \frac \epsilon 2.$$ For each $x \in (b-\delta, b)$ determine a point $t \in (b-\delta, b)$ which is so near to $b$ that \begin{align}|f(t)+g(t)| &< \frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)} \\ |g(t)| &< \frac{|g(x)|}{2}.\end{align} Since $f(t)$ and $g(t)$ tend to $0$ as $t$ tends to $b$, and since $g(x) \neq 0$ such a $t$ exists. It depends on $x$, of course. By this choice of $t$ and the Ratio Mean Value Theorem we have, for some $\theta \in (x,t),$ \begin{align*}\left\vert \frac{f'(x)}{g'(x)}-L \right\vert &= \left\vert \frac{f(x)}{g(x)}-\frac{f(x)-f(t)}{g(x)-g(t)}+\frac{f(x)-f(t)}{g(x)-g(t)} - L \right\vert \\ &\le \left\vert \frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))} \right\vert + \left\vert \frac{f'(\theta)}{g'(\theta)}-L \right\vert < \epsilon, \end{align*} which completes the proof that $f(x)/g(x) \to L$ as $x \to b.$

The part I didn't get was the last inequality

$$\left\vert \frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))} \right\vert + \left\vert \frac{f'(\theta)}{g'(\theta)}-L \right\vert < \epsilon,$$

which I'm sure relates to his constraints on $|f(t) + g(t)|$ and $g(t)$. I understood his general point about $f(t)/f(x), g(t)/g(x)$ getting arbitrarily small so that $$\frac{f(x)}{g(x)} \approx \frac{f(x)-f(t)}{g(x)-g(t)}$$ but don't really understand the finer details of the proof.

Any help is greatly appreciated. :)

Answer 1

This is a good question. I actually don't think it follows from what he has written. Take, for example, $g(t) = 1/2, f(t) = -1/2, g(x) = 1, f(x) = 1$. Then $|f(t)+g(t)| < \frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)}$ and $|g(t)| < \frac{|g(x)|}{2}$, but $|\frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))}| = |\frac{-1/2-1/2}{1/2}| = 2$.

I can't figure out what he was going for. He already said $x \in (b-\delta,b)$ implies $|\frac{f'(x)}{g'(x)}-L| < \frac{\epsilon}{2}$. He fixed an $x \in (b-\delta,b)$ and $t \in (x,b)$, so since $\theta \in (x,t)$, we know $\theta \in (b-\delta,b)$ and thus $|\frac{f'(\theta)}{g'(\theta)}-L| < \frac{\epsilon}{2}$. So we just need to show that $|\frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))}| < \frac{\epsilon}{2}$. But I don't see how the two chosen conditions on $t$ would help (indeed, the first part of my answer shows that more is needed).

Answer 2

You wrote the key inequality in the book incorrectly: $$ |f(t)|+|g(t)|<\frac{g(x)^2\varepsilon}{4(|f(x)|+|g(x)|)}\tag{1} $$ ^{(It is $|f(t)|+|g(t)|$, not $|f(t)+g(t)|$ on the left.)}

By inequality (1), one has $$ |g(x)f(t)-f(x)g(t)|\leq (|f(t)|+|g(t)|)(|f(x)|+|g(x)|)<\frac{g(x)^2\varepsilon}{4}. $$ Hence, $$ \left|\frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))}\right|\leq \left|\frac{g(x)^2\varepsilon}{4g(x)(g(x)-g(t))}\right| =\left|\frac{\varepsilon}{4(1-g(t)/g(x))}\right|. $$

But the triangle inequality implies that (where we use $|g(t)|<|g(x)|/2$) $$ |1-g(t)/g(x)|\geq 1-|g(t)/g(x)|\geq 1-\frac12=1/2. $$ The desired estimate follows: $$ \left|\frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))}\right|\leq \frac{\epsilon}{2}. $$

The other half of the estimate $\left|\frac{f'(\theta)}{g'(\theta)}-L\right|<\frac{\varepsilon}{2}$ comes from the observation that $$ \forall x\in(b-\delta,b)\quad \left|\frac{f'(x)}{g'(x)}-L\right|<\frac{\varepsilon}{2}. $$

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Here is the original excerpt in Pugh's book (2nd edition):

Answer 3

Ugh.

(1). Notice that the last line of the proof employs the Second Mean Value Theorem: If $x<t$ and $f,g$ are differentiable on an open interval containing $x$ and $t,$ and if $g'\ne 0$ on $(x,t)$ then $\frac {f(x)-f(t)}{g(x)-g(t)}=\frac {f'(\theta)}{g'(\theta)}$ for some $\theta\in (x,t).$

(2). Since $L=\lim_{x\to b^-}\frac {f'(x)}{g'(x)}$ exists, there exists $c\in (a,b)$ such that $g'\ne 0$ on $[c,b).$

Otherwise there would be values of $x$ in $(a,b)$ arbitrarily close to $b$ for which $g'(x)=0$ and hence for which $f'(x)/g'(x)$ would not exist, but then $\lim_{x\to b^-}f'(x)/g'(x)$ would not exist.

(3). With $c$ as in (1) there exists $d\in [c,b)$ such that $g\ne 0$ on $[d,b).$

Otherwise there would exist $x_1,x_2 \in [c,b)$ with $x_1< x_2$ and $g(x_1)=g(x_2)=0$, but by the First Mean Value Theorem $0=\frac {g(x_1)-g(x_2)}{x_1-x_2}=g'(x_3)$ for some $x_3\in (x_1,x_2)\subset [c,b),$ contrary to (2).

(4). With $c,d$ as in (2) and (3): Given $\epsilon >0,$ take $e\in [d,b)$ such that $y\in [e,b)\implies |L-\frac {f'(y)}{g'(y)}|<\epsilon/2.$

Then for $e\le x<t<b,$ by the Second Mean Value Theorem there exists $y\in (x,t)\subset [e,b)$ with $|L-\frac {f(x)-f(t)}{g(x)-g(t)}|=|L-\frac {f'(y)}{g'(y)}|<\epsilon/2$.

(5). The Q is now , for a given $x\in [e,b),$ whether there is some (any) $t\in (x,b)$ such that $|\frac {f(x)}{g(x)}-\frac {f(x)-f(t)}{g(x)-g(t)}|<\epsilon/2.$

With $x$ fixed and $t\to b^-$ we have $\lim_{t\to b^-} f(x)-f(t)=f(x)$ and $\lim_{t\to b^-}g(x)-g(t)=g(x)\ne 0.$ So $\lim_{t\to b^-}\frac {f(x)-f(t)}{g(x)-g(t)}=\frac {f(x)}{g(x)}.$ So any $t\in (x,b)$ with $t$ sufficiently close to $b$ will work.

We conclude that, given $\epsilon>0,$ there exists $e\in (a,b)$ such that $x\in [e,b)\implies |L-f(x)/g(x)|<\epsilon.$

Answer 4

It may come from \begin{align}|f(t)|+|g(t)| &< \frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)} \\ |g(t)| &< \frac{|g(x)|}{2}.\end{align}

Here I change $|f(t)+g(t)|$ to $|f(t)|+|g(t)|$ in the first inequality. Check if you typed wrongly or that may be a typo of the book.

Since $|g(t)| < \frac{|g(x)|}{2}$ we have $|g(x) - g(t)| > \frac{|g(x)|}{2}$. Therefore $$\left\vert \frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))} \right\vert \le \frac{|g(x)f(t)-f(x)g(t)|}{\frac{g^2(x)}{2}} \le \frac{|g(x)||f(t)| + |g(x)||g(t)| }{\frac{g^2(x)}{2}} \le \frac{(|g(t)|+|f(t)|)(|g(x)|+|f(x)|)}{\frac{g^2(x)}{2}} < \frac{\epsilon}{2}$$

The last inequality is just $|f(t)|+|g(t)| <\frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)} $. Then we've done.