# Proof of L'Hospitals Rule

No matter where I would look it would seem that L'Hospital's Rule has a strange proof-given that they teach it in high school, it seems troublesome that I can't find a solid proof at that level of knowledge. Does anyone have a proof that is fairly basic? Or does proving it simply require higher math?

• I don't recall it being proved in high school, and I've not seen a proof until looking it up on Wikipedia just now. – Ian Coley Sep 26 '13 at 4:43
• you can prove it via generalized Mean Value Theorem IRRC. – Santosh Linkha Sep 26 '13 at 4:45

A related question is how would you discover L'Hopital's rule. Assuming that $f(x) = g(x) = 0$, one might notice that \begin{align} \frac{f(x + h)}{g(x+h)} &= \frac{\frac{f(x+h) - f(x)}{h}}{\frac{g(x+h)-g(x)}{h}} \\ &\approx \frac{f'(x)}{g'(x)} \end{align} which suggests L'Hopital's rule. This could be turned into a rigorous proof, but we'd need to make some unnecessarily restrictive assumptions.

Here is a version of L'Hopital's rule with a simple proof: Assume $f$ and $g$ are differentiable at $x$ and $g'(x) \neq 0$, and that $f(x) = g(x) = 0$. Then $$\lim_{h \to 0} \frac{f(x+h)}{g(x+h)} = \frac{f'(x)}{g'(x)}.$$

Proving a less restrictive version of L'Hopital's rule requires a less obvious argument.

One can prove it using linear approximation:

Recall the definition of the derivative, $f'(x)$: $$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ which we can write as $$f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h)$$ where $\eta(h)$ is a function such that $\lim_{h\rightarrow 0} \eta(h)=0$ (and is continuos). Then in the above displayed equation multiply through by $h$, which gives (after rearrangment and switching $\eta(h)$ with $-\eta(h)$) $$f(x+h) = f(x) +f'(x)h+h\cdot \eta(h).$$ Now use L'Hopitals rule for the limit of $f(x)/g(x)$ when $x$ goes to $x_0$, so we have $f(x_0)=g(x_0)=0$. Then by using linear approximation as above in both numerator and denominator (and writing the $\eta$-function for $g$ as $\epsilon$), we get $$\frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)}$$ and now, in both numerator and denominator, there is a common factor $h$ we can cancel. Taking the limit as $h \rightarrow 0$ now gives the result.

• Very elegant proof -- I've been re-working through Rudin's proof of L'Hospital's Rule in $Principles~of~Mathematical~Analysis$, 3rd Ed., and began looking for help with it (I included the silent "s" since this is the original spelling in the late 17th century and it is included in Rudin's book). Where you say, "Now $use$ L'Hopitals rule for the limit...," are you saying to incorporate the assumptions of the theorem, where $\lim\limits_{x\rightarrow x_{0}}\dfrac{f(x)}{g(x)}=\dfrac{f(x_{0})}{g(x_{0})}=\dfrac{0}{0}$? I just wanted to make sure of this part since I very much like this proof. – Procore May 22 '17 at 21:32

Most proofs of L'Hôpital's rule requires Cauchy's mean value theorem. If the reader is familiar with that theorem and its applications, then the proof of L'Hôpital's rule is not that hard. If the use of Cauchy's theorem is the strangeness you feel, then there may not be a way around it. The following line of thought might make you feel better about accepting it, though.

Many of the "well-known" facts used in applications of differential calculus need a use of the mean-value theorem for a rigorous proof. For example,

1. If $f: (a,b) \rightarrow \mathbb{R}$ is differentiable and $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant

2. If the derivative of a function $f$ is everywhere strictly positive, then $f$ is a strictly increasing function.

3. The derivative of a differentiable function vanishes at maxima and minima.

These facts are easier than L'Hôpital's rule, and they require the ordinary mean value theorem. So it is not a stretch to accept that L'Hôpital's rule, which is a more sophisticated theorem, require a more sophisticated mean value theorem: Cauchy's mean value theorem.

• I think #1 and #2 boil down to the difficulty of proving "If the derivative of a function $f$ is everywhere $\hspace{.61 in}$ non-negative, then $f$ is non-decreasing". $\:$ I also think that #3 can easily be proven without using the MVT. $\hspace{.27 in}$ – user57159 Sep 26 '13 at 5:35
• @RickyDemer: Sources please, if you have them. – user96815 Sep 26 '13 at 7:44
• My sources are the theorems and proofs in this answer. $\;$ – user57159 Sep 26 '13 at 10:21
• @RickyDemer: Thanks. I will look into it. – user96815 Oct 3 '13 at 17:11

Typically when they teach L'Hopital's Rule in school they just teach it algorithmically, that is just how to apply it, without the proof. This is very similar to the way calculus in general is taught in most schools, i.e., just as a bunch of techniques, no proofs or justifications. A proper proof of L'Hopital's Theorem is not terribly difficult but it certainly requires a careful development of the basic language of calculus, namely the formal concept of limits and their basic theory.

A baby version of the rule can be stated and proved for power series. This can be taken as motivation for the general rule.

This is in response to Doldrums' request, but is way too long for a comment.

Theorem:

If the derivative of a function $\hspace{.04 in}f$ is everywhere non-negative, then $\hspace{.04 in}f$ is non-decreasing.
$\implies$
If $\hspace{.04 in}f$ is differentiable and $\:\hspace{.04 in}f'(x) = 0\:$ for all $\;\; x\: \in \: (a,b) \;\;$, $\;\;$ then $\hspace{.04 in}f$ is constant.

Proof:

Suppose $\hspace{.04 in}f$ is differentiable and $\:\hspace{.04 in}f'(x) = 0\:$ for all $\:\: x\: \in \: (a,b) \;\;$.
Let $\;\; g \: : \: (a,b) \to \mathbb{R} \;\;$ be given by $\: g(x) = -(\hspace{.05 in}f(x)) \;$.
For all $\:\: x\: \in \: (a,b) \:\:$,
$g'(x) \: = \: (g(x))' \: = \: (-(\hspace{.05 in}f(x)))' \: = \: -((\hspace{.05 in}f(x))') \: = \: -(\hspace{.05 in}f'(x)) = -0 = 0 \;\;\;$.
For all elements $x$ and $y$ of $(a,b)$, $\;\;$[$\;\; \hspace{.04 in}f(x) \leq \hspace{.04 in}f(y) \: \text{ and } \: g(x) \leq g(y)$
$\implies \;\;\hspace{.04 in}f(x) \leq \hspace{.04 in}f(y) \: \text{ and } \: -\hspace{-0.04 in}(\hspace{.05 in}f(x)) = g(x) \leq g(\hspace{.03 in}y) = -(\hspace{.05 in}f(\hspace{.03 in}y))$
$\implies \;\; \hspace{.04 in}f(x) \leq \hspace{.04 in}f(\hspace{.03 in}y) \: \text{ and } \: \hspace{.04 in}f(\hspace{.03 in}y) \leq \hspace{.04 in}f(x) \;\; \implies \;\; \hspace{.04 in}f(x) = \hspace{.04 in}f(\hspace{.03 in}y) \;\;$]$\;\;$.
If $\hspace{.04 in}f$ and $g$ are non-decreasing, then $\hspace{.04 in}f$ is constant.
If the first sentence of the theorem's (main) implication holds,
then the second sentence of that theorem holds. $\:$ Therefore the theorem holds.

Theorem: $\;\;\;$ If $\:\operatorname{Dom}(\hspace{.05 in}f\hspace{.02 in})\:$ is an interval then

If the derivative of a function $\hspace{.04 in}f$ is everywhere non-negative, then $\hspace{.04 in}f$ is non-decreasing.
$\implies$
If the derivative of a function $\hspace{.04 in}f$ is everywhere strictly positive,
then $\hspace{.04 in}f$ is a strictly increasing function.

.

Proof:

Assume that the first sentence of the theorem's arrow holds, and then suppose towards a contradiction that the derivative of $\hspace{.04 in}f$ is everywhere strictly positive, and that the inputs $x$ and $y$ such that $\: x< y\:$ and $\: \hspace{.04 in}f(x) \leq \hspace{.04 in}f(\hspace{.03 in}y) \;$. $\;\;\;$ For all elements $z$ of $[x,y]$, $\: \hspace{.04 in}f(x) \leq \hspace{.04 in}f(z) \leq \hspace{.04 in}f(\hspace{.03 in}y) = \hspace{.04 in}f(x) \;$.
$\hspace{.04 in}f$ is constant on $[x,y]$. $\;\;\;$ For $\: w = \frac{x+z}2 \:$ and $h$ such that $\: 0 < |h| < \frac{y-x}2 \:$, $\:$ $\;\; w+h \: \in \: [x,y] \;\;\;$.
For $\: w = \frac{x+z}2 \:$ and $h$ such that $\: 0 < |h| < \frac{y-x}2 \:$, $\:$ $\: \frac{\hspace{.04 in}f(w+h)-\hspace{.04 in}f(w)}h = \frac{\hspace{.04 in}f(w+0)-\hspace{.04 in}f(w+0)}h = \frac{0-0}h = \frac0h = 0 \;\;$.
Since $\: x< y\:$, $\: 0 < \frac{y-x}2 \:$, $\:$ so $\;\; \hspace{.04 in}f'(w) \; = \; \displaystyle\lim_{h\to 0} \frac{\hspace{.04 in}f(w+h)-\hspace{.04 in}f(w)}h \; = \; \displaystyle\lim_{h\to 0} \: 0 \; = \; 0 \;\;$,
contradicting the assumption of the second sentence of the theorem's arrow.
This contradiction shows that the second sentence of the theorem's arrow follows
from the assumption at the beginning of this proof. $\:$ Therefore the theorem holds.

Theorem:

If $x$ is in the interior of $\operatorname{Dom}(\hspace{.05 in}f\hspace{.02 in})$ and $x$ is a local weak
extremum of $\hspace{.04 in}f$ and $\hspace{.04 in}f'(x)$ exists, then $\: \hspace{.04 in}f'(x) = 0 \;$.

Proof:

Suppose $\epsilon$ is such that $\:0<\epsilon \:$ and $\: (x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon) \subseteq \operatorname{Dom}(\hspace{.05 in}f\hspace{.02 in}) \:$ and $x$ weakly
maximizes $\hspace{.04 in}f$ on the interval $\:(x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon)\;$. $\;\;\;$ For all $h$, $\: \big|\hspace{-0.04 in}\pm\hspace{-0.04 in}|h|\big| = |h| \;$.
For all $h$, if $\: 0<|\hspace{.02 in}h\hspace{.01 in}|<\epsilon \:$ then $\:h\neq 0\:$ and $\;\; x\hspace{-0.02 in}\pm\hspace{-0.01 in}|h| \: \in \: (x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon) \subseteq \operatorname{Dom}(\hspace{.05 in}f\hspace{.02 in}) \;\;$.
For all $h$, if $\: 0<|\hspace{.02 in}h\hspace{.01 in}|<\epsilon \:$ then $\:\: \frac{\hspace{.04 in}f(x+|h|)-f(x)}{|h|} \leq \frac{\hspace{.04 in}f(x)-\hspace{.04 in}f(x)}{|h|} = \frac{0-0}{|h|} = \frac0{|h|} = 0 \;\;$.
For all $h$, if $\: 0<-|\hspace{.02 in}h\hspace{.01 in}|<\epsilon \:$ then $0 = \frac0{|h|} = \frac{0-0}{|h|} = \frac{\hspace{.04 in}f(x)-\hspace{.04 in}f(x)}{|h|} = \frac{\hspace{.04 in}f(x)-\hspace{.04 in}f(x+(-|h|))}{|h|} \leq \frac{-(\hspace{.04 in}f(x+(-|h|))-f(x))}{|h|} = \frac{--(\hspace{.04 in}f(x+(-|h|))-f(x))}{-|h|} = \frac{\hspace{.04 in}f(x+(-|h|))-f(x)}{-|h|} \;\;$.
For all $h$, if $\: 0 < |h| < \epsilon \:$ then $\: \frac{\hspace{.04 in}f(x+|h|)-f(x)}{|h|} \leq 0\leq \frac{\hspace{.04 in}f(x+(-|h|))-f(x)}{-|h|} \;$.
If $\; \displaystyle\lim_{h\to 0} \frac{\hspace{.04 in}f(x+h)-f(x)}h \;$ exists then $\;\; \displaystyle\lim_{h\to 0} \frac{\hspace{.04 in}f(x+h)-f(x)}h \; = \; 0 \;\;\;$.
If $\hspace{.04 in}f'(x)$ exists then $\: \hspace{.04 in}f'(x) = 0 \;$.
Now suppose $\epsilon$ is such that $\:0<\epsilon \:$ and $x$ weakly minimizes $\hspace{.04 in}f$ on the interval $\:(x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon)\;$.
Let $g$ be given by $\:g(y) = -(\hspace{.05 in}f(y))\;$. $\;\;\;$ For all $\: y\in (x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon) \:$,
$-(g(x)) \; = \; --(\hspace{.05 in}f(x)) \; = \; \hspace{.04 in}f(x) \; \leq \; \hspace{.04 in}f(y) \; = \; --(\hspace{.05 in}f(y)) \; = \; -(g(y)) \;\;\;$.
For all $\: y\in (x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon) \:$, $\:$ $\: g(y) \leq g(x) \;$. $\;\;\;$ $x$ weakly maximizes $g$ on $\:(x\hspace{-0.04 in}-\hspace{-0.04 in}\epsilon,x\hspace{-0.04 in}+\hspace{-0.04 in}\epsilon)\;$.
By the earlier part of this proof, if $g'(x)$ exists then $\:g'(x) = 0 \;$. $\;\;\;$ If $\hspace{.04 in}f'(x)$ exists then
$\hspace{.04 in}f'(x) \: = \: (\hspace{.05 in}f(x))' \: = \: -\hspace{-0.04 in}-\hspace{-0.04 in}((\hspace{.05 in}f(x))') \: = \: -((-(\hspace{.05 in}f(x)))') \: = \: -((g(x))') \: = \: -(g'(x)) \: = \: -0 \: = \: 0 \;\;\;$.
In both cases, one obtained that "if $\hspace{.04 in}f'(x)$ exists then $\:\hspace{.04 in}f'(x) = 0\:$". $\;\;$ Therefore the theorem holds.