# Clarification of L'Hopital Proof Pugh

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 $$g(x),g'(x) \neq 0.$$

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. :)

• It's the MVT followed by the triangle inequality. – Matematleta Jun 22 '19 at 23:38
• There is a typo in the first expression in the last displayed set. It should be $|f(x)/g(x)-L|,$ not $|f'(x)/g'(x)-L|$. – DanielWainfleet Oct 18 '19 at 9:01
• I must say this is the most complicated proof I have seen for L'Hospital's Rule. The proof is easily handled by defining $f(b) =g(b) =0$ and applying Cauchy MVT on $f, g$ on interval $[x, b]$. Why all this drama of choosing $t$? – Paramanand Singh Oct 18 '19 at 14:28
• @ParamanandSingh more intuitive – mathworker21 Oct 19 '19 at 22:03
• @ParamanandSingh I think one reason is that this proof can be easily extended to the case where the interval is (a,inf) – Amirh.Kp Nov 7 '19 at 13:36

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 $$\forall x\in(b-\delta,b)\quad \left|\frac{f'(x)}{g'(x)}-L\right|<\frac{\varepsilon}{2}.$$

Here is the original excerpt in Pugh's book (2nd edition): • how did you get $|g(x)f(t)-f(x)g(t)|\leq (|f(t)|+|g(t)|)(|f(x)|+|g(x)|)$? – Donlans Donlans Oct 18 '19 at 20:02
• @DonlansDonlans: The right hand side is at least $|g(x)||f(t)|+|f(x)||g(t)|$ (by simply ignoring the other two terms). – Jack Oct 18 '19 at 20:06
• @DonlansDonlans then one can apply the triangle inequality for the left hand side. – Jack Oct 18 '19 at 20:22

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).

• I appreciate your insightful contribution to this older Question. – hardmath Oct 23 '19 at 17:02
• Hi @mathworker21, I have asked a related question math.stackexchange.com/questions/3410301/…, but have received no answer. Could you please have a look at it? Thank you for your help! – Akira Oct 27 '19 at 7:06

Ugh.

(1). Notice that the last line of the proof employs the Second Mean Value Theorem: If $$x and $$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 $$e\le x by the Second Mean Value Theorem there exists $$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 $$x\in [e,b)\implies |L-f(x)/g(x)|<\epsilon.$$

• Parts (2) and (3) are preliminary work. We have to establish that $f(x)/g(x)$ exists (i.e. $g(x)\ne 0$) for all $x$ sufficiently close to $b.$ – DanielWainfleet Oct 18 '19 at 11:01
• Note that in (4) and (5) the denominator $g(x)-g(t)\ne 0.$ Otherwise for some $z\in (x,t) \subset [c,b)$ we would have $0=(g(x)-g(t))/(x-t)=g'(z),$ contrary to (2). – DanielWainfleet Oct 18 '19 at 11:21
• Thanks for taking the time to give a rigorous treatment. – hardmath Oct 23 '19 at 17:04
• I knew I had seen a simple proof when I was young, in the Jurassic. If the author hadn't invoked the 2nd mean value theorem I likely wouldn't have recalled it. L'Hopital's theorem is one of those that we use a lot while forgetting the proof. – DanielWainfleet Oct 31 '19 at 16:58

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.

• i just looked at book. OP typed wrongly – mathworker21 Oct 18 '19 at 4:05
• @mathworker21: It could also be that it was an error in one edition of the book that was corrected in a later edition. – celtschk Oct 20 '19 at 21:33
• @celtschk ah, yes, good point. "looked at book" is not well-defined – mathworker21 Oct 20 '19 at 21:34