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. :)

  • 1
    $\begingroup$ It's the MVT followed by the triangle inequality. $\endgroup$ Jun 22, 2019 at 23:38
  • 1
    $\begingroup$ There is a typo in the first expression in the last displayed set. It should be $|f(x)/g(x)-L|,$ not $|f'(x)/g'(x)-L|$. $\endgroup$ Oct 18, 2019 at 9:01
  • 6
    $\begingroup$ 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 $t$? $\endgroup$
    – Paramanand Singh
    Oct 18, 2019 at 14:28
  • $\begingroup$ @ParamanandSingh more intuitive $\endgroup$ Oct 19, 2019 at 22:03
  • $\begingroup$ @ParamanandSingh I think one reason is that this proof can be easily extended to the case where the interval is (a,inf) $\endgroup$
    – Amirh.Kp
    Nov 7, 2019 at 13:36

4 Answers 4


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).

  • $\begingroup$ I appreciate your insightful contribution to this older Question. $\endgroup$
    – hardmath
    Oct 23, 2019 at 17:02
  • $\begingroup$ 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! $\endgroup$
    – Akira
    Oct 27, 2019 at 7:06

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):

enter image description here

  • $\begingroup$ how did you get $|g(x)f(t)-f(x)g(t)|\leq (|f(t)|+|g(t)|)(|f(x)|+|g(x)|)$? $\endgroup$ Oct 18, 2019 at 20:02
  • $\begingroup$ @DonlansDonlans: The right hand side is at least $|g(x)||f(t)|+|f(x)||g(t)|$ (by simply ignoring the other two terms). $\endgroup$
    – user9464
    Oct 18, 2019 at 20:06
  • $\begingroup$ @DonlansDonlans then one can apply the triangle inequality for the left hand side. $\endgroup$
    – user9464
    Oct 18, 2019 at 20:22


(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 $x\in [e,b)\implies |L-f(x)/g(x)|<\epsilon.$

  • 1
    $\begingroup$ 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 $b.$ $\endgroup$ Oct 18, 2019 at 11:01
  • $\begingroup$ 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). $\endgroup$ Oct 18, 2019 at 11:21
  • $\begingroup$ Thanks for taking the time to give a rigorous treatment. $\endgroup$
    – hardmath
    Oct 23, 2019 at 17:04
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 31, 2019 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.

  • $\begingroup$ i just looked at book. OP typed wrongly $\endgroup$ Oct 18, 2019 at 4:05
  • 2
    $\begingroup$ @mathworker21: It could also be that it was an error in one edition of the book that was corrected in a later edition. $\endgroup$
    – celtschk
    Oct 20, 2019 at 21:33
  • $\begingroup$ @celtschk ah, yes, good point. "looked at book" is not well-defined $\endgroup$ Oct 20, 2019 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.