Why does L'Hospital's rule require the limit to exist? About the proof.

Question

Proof of L'Hospital's rule (only special case this time):

Let's assume, that $$ f(a)=g(a)=0 .$$

Using the MVT, we get $$ \frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\zeta)}{g'(\zeta)}, \ \mathrm{where} \ \zeta \in (a, x) \ \mathrm{or} \ \zeta \in (x, a) $$ and thus $$ \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f'(\zeta)}{g'(\zeta)}=\lim_{\zeta \rightarrow a} \frac{f'(\zeta)}{g'(\zeta)}=\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}. $$

In literature, it's always assumed that the limit $$ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} $$ exists. Why is that and how is it shown in the proof? According to the proof above, shouldn't the limit of f/g be equal to the limit of f'/g' always? If lim f'/g' is not defined, then lim f/g shouldn't be defined either.

Obviously this isn't true, because there are examples of functions f and g s.t. lim f/g exists even though lim f'/g' does not exist.

Answer 1

L'Hôpital's rule is an "if" statement, not an "if and only if" statement. It tell us that if

$\displaystyle \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$

exists then

$\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$

but it says nothing about the case when $\displaystyle \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$ does not exist. In that case $\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}$ may or may not exist - as you say, there are examples of each possibility.

If $\displaystyle \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$ does not exist then, although we may know that

$\displaystyle \frac {f(x)}{g(x)} = \frac {f'(\zeta_1)}{g'(\zeta_2)}$

for some $\zeta_1, \zeta_2 \in (x,a)$, the right hand side does not tend to a specific value as $x \rightarrow a$, so we cannot conclude anything about the behaviour of the left hand side.

Answer 2

L'Hospital's rule "requires" the limit to exist, because if it does not, you are stuck and the rule is useless. In particular, the functions $f,g$ might be non-differentiable.

Answer 3

I think the confusion comes from the particular version of L'Hosital one is using. Here is a version of L'Hospital that explicitly requires the existence of the limit the ratio of derivatives. I will also provide an example where this fails. I hope this helps.

Theorem: Suppose $f,g$ are differentiable functions in an interval $(a,b)$. If

1. $\lim_{x\rightarrow a+}f(x)=0=\lim_{x\rightarrow a+}g(x)$,
2. $g'(x)\neq0$ in $(a,b)$, and
3. $\lim_{x\rightarrow a+}\frac{f'(x)}{g'(x)}$ exists and has value $L$ (here $L$ is either a real number, $\infty$ or $-\infty$)

then $\lim_{x\rightarrow a+}\frac{f(x)}{g(x)}$ exists and equals $L$.

Notes:

• A similar result holds for $x\rightarrow b-$.
• Similar versions exist when in (1), $0$ is replace by $\pm\infty$.

---

Here is a short proof for $a$ and $L$ are finite. By condition (1) we can extend $f$ and $g$ to $[a,b)$ be setting $f(a)=0=g(a)$. Given $\varepsilon>0$ there is $x_\varepsilon\in (a,b)$ such that $$ \Big|\frac{f'(x)}{g'(x)}-L\Big|<\varepsilon, \qquad\text{for all}\quad a<x\leq x_\varepsilon $$ By the mean value theorem (the generalized version), for each $a<x<x_\varepsilon$, $$ \frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(c_x)}{g'(c_x)}\qquad\text{for some}\quad a<c_x<x. $$ Hence, for all $a<x<x_\varepsilon$ $$ \Big|\frac{f(x)}{g(x)} -L\Big|=\Big|\frac{f'(c_x)}{g'(c_x)} -L\Big|<\varepsilon $$

The case where $L$ is not finite is handled similarly. When $a=-\infty$ ($b=\infty$), a slight modification of the proof above works.

---

Observations: Of all the assumptions made in the statement of L'Hospital theorem, assumption (3) is the source of many misconceptions among students. That is because while (1) and (2) are easy to check, (3) is typically not a priory verifiable. Assumption (3) is a sort of the clarivoyant condition.

• It is important to understand what L'Hospital actually says: Under condition (1) and (2), the existence of $\lim_{x\rightarrow a+}\frac{f'(x)}{g'(x)}$ implies the existence of $\lim_{x\rightarrow a+}\frac{f(x)}{g(x)}$, but no the other way around (see examples 1, 3).

• Under conditions (1) and (2), the existence of $\lim_{x\rightarrow a+}\frac{f(x)}{g(x)}$ does not imply the existence of $\lim_{x\rightarrow a+}\frac{f'(x)}{g'(x)}$. Certainly, by looking at the proof outlined above, the existence of the former implies the existence of the latter along a subset of values ($c_x$) approaching $a+$, but that is not the same as saying that the limit $\lim_{x\rightarrow a+}f'(x)/g'(x)$ (unconstrained on how one approaches $a+$) exists (see example 3).

• The lack of a priori information on condition (3) implies that when one applies the procedure of replacing the estimation of the limit of $f/g$ by that of estimating the the limit of ratio $f'/g'$, the outcome may be uncertain and inconclusive (going around loops).

• The value of $g'(a)$, in the case where $f$ and $g$ can be extended continuously to $[a,b)$ and and the right derivative $g'(a+)$ exists, is irrelevant. It may not even be defined.

---

Some examples

1. Here L'Hospital rule is not applicable as (3) does not hold: $$\lim_{x\rightarrow\infty}\frac{x-\sin x}{x+\sin x}=\lim_{x\rightarrow\infty}\frac{1-\frac{\sin x}{x}}{1+\frac{\sin x}{x}}=1$$ however $$\lim_{x\rightarrow\infty}\frac{\Big(x-\sin x\Big)'}{\Big(x+\sin x\Big)'}=\lim_{x\rightarrow\infty}\frac{1-\cos x}{1+\cos x}$$ does not exists, as one can see by looking at the sequence $y_n=2n\pi$ and $z_n=\frac{\pi}{2}+2\pi n$.

2. Here L'Hospital rule have us going in an infinite loop (gives us back the problem we started with) $$\lim_{x\rightarrow\infty}\frac{x}{\sqrt{1+x^2}}=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}}=1$$ but $$\lim_{x\rightarrow\infty}\frac{\big(x\big)'}{\big(\sqrt{1+x^2}\big)'}=\lim_{x\rightarrow\infty}\frac{\sqrt{1+x^2}}{x}=\lim_{x\rightarrow\infty}\frac{1}{\frac{x}{\sqrt{1+x^2}}}$$ So in a way you go back to your original problem.

3. Here the assumption (3) of L'Hospital does not hold but one may not see that $$ \lim_{x\rightarrow0}\frac{x^2\sin(x^{-1})}{\sin x}=\lim_{x\rightarrow0}\frac{x\sin(x^{-1})}{\frac{\sin x}{x}}=0$$ sine $|x\sin(x^{-1})|\leq|x|\xrightarrow{x\rightarrow0}$ and $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$. However

$$ \lim_{x\rightarrow0}\frac{\big(x^2\sin(x^{-1})\big)'}{(\sin x)'}= \lim_{x\rightarrow0}\frac{2x\sin(x^{-1}) -\cos(x^{-1})}{\cos x} $$ But $\lim_{x\rightarrow0}\cos(x^{-1})$ does not exists, as one can check by looking at the sequences $y_n=\frac{1}{2n\pi}$ and $x_n=\frac{1}{\frac{\pi}{2}+2\pi n}$