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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. :)

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    $\begingroup$ It's the MVT followed by the triangle inequality. $\endgroup$ – Matematleta Jun 22 at 23:38
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    $\begingroup$ There is a typo in the first expression in the last displayed set. It should be $|f(x)/g(x)-L|,$ not $|f'(x)/g'(x)-L|$. $\endgroup$ – DanielWainfleet Oct 18 at 9:01
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    $\begingroup$ 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 at 14:28
  • $\begingroup$ @ParamanandSingh more intuitive $\endgroup$ – mathworker21 Oct 19 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 at 13:36
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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

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  • $\begingroup$ how did you get $|g(x)f(t)-f(x)g(t)|\leq (|f(t)|+|g(t)|)(|f(x)|+|g(x)|)$? $\endgroup$ – Donlans Donlans Oct 18 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). $\endgroup$ – Jack Oct 18 at 20:06
  • $\begingroup$ @DonlansDonlans then one can apply the triangle inequality for the left hand side. $\endgroup$ – Jack Oct 18 at 20:22
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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).

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  • $\begingroup$ I appreciate your insightful contribution to this older Question. $\endgroup$ – hardmath Oct 23 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 at 7:06
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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 $x\in [e,b)\implies |L-f(x)/g(x)|<\epsilon.$

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    $\begingroup$ 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 $b.$ $\endgroup$ – DanielWainfleet Oct 18 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). $\endgroup$ – DanielWainfleet Oct 18 at 11:21
  • $\begingroup$ Thanks for taking the time to give a rigorous treatment. $\endgroup$ – hardmath Oct 23 at 17:04
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    $\begingroup$ 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. $\endgroup$ – DanielWainfleet Oct 31 at 16:58
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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.

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  • $\begingroup$ i just looked at book. OP typed wrongly $\endgroup$ – mathworker21 Oct 18 at 4:05
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    $\begingroup$ @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 at 21:33
  • $\begingroup$ @celtschk ah, yes, good point. "looked at book" is not well-defined $\endgroup$ – mathworker21 Oct 20 at 21:34

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