L'Hôpital's Rule and Indeterminate Forms

Subsection 3.7.1 L'Hôpital's Rule and Indeterminate Forms

Let us return to limits (Chapter 1) and see how we can use derivatives to simplify certain families of limits called indeterminate forms. We know, from Theorem 1.4.3 on the arithmetic of limits, that if

\begin{align*} \lim_{x\rightarrow a}f(x) &= F & \lim_{x\rightarrow a}g(x) &= G\\ \end{align*}

and $G\ne 0\text{,}$ then

\begin{align*} \lim_{x\rightarrow a}\frac{f(x)}{g(x)} &= \frac{F}{G} \end{align*}

The requirement that $G\ne 0$ is critical — we explored this in Example 1.4.7. Please reread that example.

Of course ¹Now it is not so surprising, but perhaps back when we started limits, this was not so obvious. it is not surprising that if $F\ne 0$ and $G= 0\text{,}$ then

\begin{align*} \lim_{x\to a}\frac{f(x)}{g(x)} &= DNE \end{align*}

and if $F=0$ but $G\neq 0$ then

\begin{align*} \lim_{x\to a}\frac{f(x)}{g(x)} &= 0 \end{align*}

However when both $F,G=0$ then, as we saw in Example 1.4.7, almost anything can happen

\begin{align*} f(x)&=x & g(x)&=x^2 & \lim_{x\to0} \frac{x}{x^2} &= \lim_{x\to0} \frac{1}{x} = DNE\\ f(x)&=x^2 & g(x)&=x & \lim_{x\to0} \frac{x^2}{x} &= \lim_{x\to0} x = 0\\ f(x)&=x & g(x)&=x & \lim_{x\to0} \frac{x}{x} &= \lim_{x\to0} 1 = 1\\ f(x)&=7x^2 & g(x)&=3x^2 & \lim_{x\to0} \frac{7x^2}{3x^2} &= \lim_{x\to0} \frac{7}{3} = \frac{7}{3} \end{align*}

Indeed after exploring Example 1.4.12 and 1.4.14 we gave ourselves the rule of thumb that if we found $0/0\text{,}$ then there must be something that cancels.

Because the limit that results from these $0/0$ situations is not immediately obvious, but also leads to some interesting mathematics, we should give it a name.

Definition 3.7.1 First indeterminate forms

Let $a \in \mathbb{R}$ and let $f(x)$ and $g(x)$ be functions. If

\begin{align*} \lim_{x\to a} f(x) &= 0 & \text{and } && \lim_{x\to a} g(x) &= 0 \end{align*}

then the limit

\begin{gather*} \lim_{x\to a} \dfrac{f(x)}{g(x)} \end{gather*}

is called a $\frac{0}{0}$ indeterminate form.

There are quite a number of mathematical tools for evaluating such indeterminate forms — Taylor series for example. A simpler method, which works in quite a few cases, is L'Hôpital's rule ²Named for the 17th century mathematician, Guillaume de l'Hôpital, who published the first textbook on differential calculus. The eponymous rule appears in that text, but is believed to have been developed by Johann Bernoulli. The book was the source of some controversy since it contained many results by Bernoulli, which l'Hôpital acknowledged in the preface, but Bernoulli felt that l'Hôpital got undue credit.

\begin{equation*} \mbox{ } \end{equation*}

Note that around that time l'Hôpital's name was commonly spelled l'Hospital, but the spelling of silent s in French was changed subsequently; many texts spell his name l'Hospital. If you find yourself in Paris, you can hunt along Boulevard de l'Hôpital for older street signs carved into the sides of buildings which spell it “l'Hospital” — though arguably there are better things to do there. .

Theorem 3.7.2 L'Hôpital's Rule

Let $a\in\mathbb{R}$ and assume that

\begin{align*} \lim_{x\to a}f(x) &= \lim_{x\to a} g(x) = 0 \end{align*}

Then

if $f'(a)$ and $g'(a)$ exist and $g'(a)\ne 0\text{,}$ then
\begin{align*} \lim_{x\to a} \frac{f(x)}{g(x)} &=\frac{f'(a)}{g'(a)}, \end{align*}
while, if $f'(x)$ and $g'(x)$ exist, with $g'(x)$ nonzero, on an open interval that contains $a\text{,}$ except possibly at $a$ itself, and if the limit
\begin{gather*} \lim\limits_{x\rightarrow a}\frac{f'(x)}{g'(x)} \text{ exists or is $+\infty$ or is $-\infty$} \end{gather*}
then
\begin{gather*} \lim_{x\to a} \frac{f(x)}{g(x)} =\lim_{x \to a} \frac{f'(x)}{g'(x)} \end{gather*}

Proof

We only give the proof for part (a). The proof of part (b) is not very difficult, but uses the Generalised Mean–Value Theorem (Theorem 3.4.38), which is optional and most readers have not seen it.

First note that we must have $f(a)=g(a)=0\text{.}$ To see this note that since derivative $f'(a)$ exists, we know that the limit
\begin{align*} \lim_{x\to a} \frac{f(x)-f(a)}{x-a} & \text{ exists} \end{align*}
Since we know that the denominator goes to zero, we must also have that the numerator goes to zero (otherwise the limit would be undefined). Hence we must have
\begin{align*} \lim_{x\to a} (f(x)-f(a)) &= \left( \lim_{x\to a} f(x) \right) - f(a) = 0 \end{align*}
We are told that $\ds \lim_{x\to a} f(x) =0$ so we must have $f(a)=0\text{.}$ Similarly we know that $g(a)=0\text{.}$
Now consider the indeterminate form
\begin{align*} \lim_{x\to a} \frac{f(x)}{g(x)} &= \lim_{x\to a} \frac{f(x) - 0 }{g(x) - 0 } & \text{use }0=f(a)=g(a)\\ &= \lim_{x\to a} \frac{f(x) - f(a) }{g(x) - g(a) } & \text{multiply by } 1=\frac{(x-a)^{-1}}{(x-a)^{-1}}\\ &= \lim_{x\to a} \frac{f(x) - f(a) }{g(x) - g(a) }\cdot\frac{(x-a)^{-1}}{(x-a)^{-1}} & \text{rearrange}\\ &= \lim_{x\to a} \left[ \frac{ \dfrac{f(x) - f(a)}{x-a} }{ \dfrac{g(x) - g(a)}{x-a}} \right] & \text{use arithmetic of limits}\\ &= \dfrac{\ds \lim_{x\to a} \frac{f(x) - f(a)}{x-a} } {\ds \lim_{x\to a} \frac{g(x) - g(a)}{x-a}} = \dfrac{f'(a)}{g'(a)} \end{align*}
We can justify this step and apply Theorem 1.4.3, since the limits in the numerator and denominator exist, because they are just $f'(a)$ and $g'(a)\text{.}$

Subsubsection 3.7.1.1 Optional — Proof of Part (b) of l'Hôpital's Rule

To prove part (b) we must work around the possibility that $f'(a)$ and $g'(a)$ do not exist or that $f'(x)$ and $g'(x)$ are not continuous at $x=a\text{.}$ To do this, we make use of the Generalised Mean-Value Theorem (Theorem 3.4.38) that was used to prove Equation 3.4.33. We recommend you review the GMVT before proceeding.

For simplicity we consider the limit

\begin{gather*} \lim_{x\to a^+} \frac{f(x)}{g(x)} \end{gather*}

By assumption, we know that

\begin{gather*} \lim_{x\to a^+} f(x) = \lim_{x\to a^+} g(x) = 0 \end{gather*}

For simplicity, we also assume that $f(a)=g(a)=0\text{.}$ This allows us to write

\begin{align*} \frac{f(x)}{g(x)} &= \frac{f(x)-f(a)}{g(x)-g(a)} \end{align*}

which is the right form for an application of the GMVT.

By assumption $f'(x)$ and $g'(x)$ exist, with $g'(x)$ nonzero, in some open interval around $a\text{,}$ except possibly at $a$ itself. So we know that they exist, with $g'(x)\ne 0\text{,}$ in some interval $(a,b]$ with $b \gt a\text{.}$ Then the GMVT (Theorem 3.4.38) tells us that for $x\in (a,b]$

\begin{align*} \frac{f(x)}{g(x)} = \frac{f(x)-f(a)}{g(x)-g(a)} &= \frac{f'(c)}{g'(c)} \end{align*}

where $c \in (a,x)\text{.}$ As we take the limit as $x\to a\text{,}$ we also have that $c\to a\text{,}$ and so

\begin{align*} \lim_{x\to a^+}\frac{f(x)}{g(x)} &= \lim_{x\to a^+} \frac{f'(c)}{g'(c)} = \lim_{c \to a^+} \frac{f'(c)}{g'(c)} \end{align*}

as required.