4.6 L'hopital's rule

Many limits of certain combinations of functions are difficult to evaluate because they lead to what's known as“indeterminate forms.” These are expressions of the form $0/0,$ $\infty /\infty ,$ $0^0,$ $\infty -\infty ,$ $1^{\infty },$ and the like. They are precisely combinations of functions that are not covered by our limit theorems. See [link] . The very definition of the derivative itself is such a case: ${lim}_{h\to 0}(f(c+h)-f\left(c\right))=0,$ ${lim}_{h\to 0}h=0,$ and we are interested in the limit of the quotient of these two functions, which would lead us to the indeterminate form $0/0.$ The definition of the number $e$ is another example: $lim(1+1/n)=1,$ $limn=\infty ,$ and we are interested in the limit of ${(1+1/n)}^n,$ which leads to the indeterminate form $1^{\infty }.$ L'Hopital's Rule, [link] below, is our strongest tool for handling such indeterminate forms.

To begin with, here is a useful generalization of the Mean Value Theorem.

Cauchy mean value theorem

Let $f$ and $g$ be continuous real-valued functions on a closed interval $[a,b],$ suppose $g\left(a\right)\ne g\left(b\right),$ and assume that both $f$ and $g$ are differentiable on the open interval $(a,b).$ Then there exists a point $c\in (a,b)$ such that

Suppose $f$ and $g$ are differentiable real-valued functions on the bounded open interval $(a,b)$ and assume that

where $L$ is a real number. (Implicit in this hypothesis is that $g^{\text{'}}\left(x\right)\ne 0$ for $x$ in some interval $(a,a+\alpha ).$ ) Suppose further that either

or

then

Suppose first that

Observe first that, because $g^{\text{'}}\left(x\right)\ne 0$ for all $x$ in some interval $(a,a+\alpha ),$ $g^{\text{'}}\left(x\right)$ is either always positive or always negative on that interval. (This follows from part (d) of [link] .) Thereforethe function $g$ must be strictly monotonic on the interval $(a,a+\alpha ).$ Hence, since ${lim}_{x\to a+0}g\left(x\right)=0,$ we must have that $g\left(x\right)\ne 0$ on the interval $(a,a+\alpha ).$

Now, given an $ϵ>0,$ choose a positive $\delta <\alpha$ such that if $a<c<a+\delta$ then $|\frac{f^{\text{'}}\left(c\right)}{g^{\text{'}}\left(c\right)}-L|<ϵ.$ Then, for every natural number $n$ for which $1/n<\delta ,$ and every $a<x<a+\delta ,$ we have by the Cauchy Mean Value Theorem that there exists a point $c$ between $a+1/n$ and $x$ such that

Therefore, taking the limit as $n$ approaches $\infty ,$ we obtain

for all $x$ for which $a<x<a+\delta .$ This proves the theorem in this first case.

Next, suppose that

This part of the theorem is a bit more complicated to prove. First, choose a positive $\alpha$ so that $f\left(x\right)$ and $g\left(x\right)$ are both positive on the interval $(a,a+\alpha ).$ This is possible because both functions are tending to infinity as $x$ approaches $a.$ Now, given an $ϵ>0,$ choose a positive number $\beta <\alpha$ such that

for all $a<c<a+\beta .$ We express this absolute value inequality as the following pair of ordinary inequalities:

Set $y=a+\beta .$ Using the Cauchy Mean Value Theorem, and the preceding inequalities, we have that for all $a<x<y$

implying that

Dividing through by $g\left(x\right)$ and simplifying we obtain

Finally, using the hypothesis that ${lim}_{x\to a+0}g\left(x\right)=\infty ,$ and the fact that $L,ϵ,g\left(y\right),$ and $f\left(y\right)$ are all constants, choose a $\delta >0,$ with $\delta <\beta ,$ such that if $a<x<a+\delta ,$ then

and

Then, for all $a<x<a+\delta ,$ we would have

implying that

and the theorem is proved.