# How to rigorously fill in certain steps in the proof of L'Hôpital's Rule as it appears in Spivak's Calculus? [duplicate]

The following is L'Hôpital's Rule as it appears in Spivak's Calculus

Suppose that $$\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x)=0$$

and suppose also that $$\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$ exists.

Then $$\lim\limits_{x \to a} \frac{f(x)}{g(x)}$$ exists, and

$$\lim\limits_{x \to a} \frac{f(x)}{g(x)}= \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$

Spivak then provides a proof of the theorem, and at the very end writes

(Once again, the reader is invited to supply the details of this part of the argument)

The following is my attempt at supplying details of certain steps in the proof. There are two steps which I understand at an intuitive level, but not at a formal mathematical level.

The hypothesis that $$\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$ exists contains two implicit assumptions

(1) there is an interval $$(a-\delta, a+\delta)$$ such that $$f'(x)$$ and $$g'(x)$$ exist for all $$x$$ in $$(a-\delta, a+\delta)$$ except, perhaps, for $$x=a$$

(2) in this interval $$g'(x) \neq 0$$ with, once again, the possible exception of $$x=a$$

To show this, recall that $$\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}=l$$ means

$$\forall \epsilon>0\ \exists \delta>0\ \forall x,\ 0<|x-a|<\delta \implies \left | \frac{f'(x)}{g'(x)} -l \right |<\epsilon$$

Therefore, $$\frac{f'(x)}{g'(x)}$$ is defined in $$(a-\delta) \cup (a+\delta)$$, so $$f'(x)$$ and $$g'(x)$$ both exist and $$g'(x) \neq 0$$ in this interval.

Note that since this is a limit for $$x$$ approaching $$a$$ nothing is said about $$f'$$ and $$g'$$ at $$a$$.

On the other hand, $$f$$ and $$g$$ are not even assumed to be defined at $$a$$

That is, all we assumed was $$\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x)=0$$. This says something about values of $$f$$ and $$g$$ near $$a$$ but not at $$a$$.

If we define $$f(a)=g(a)=0$$ (changing the previous values of $$f(a)$$ and $$g(a)$$, if necessary), then $$f$$ and $$g$$ are continuous at $$a$$. If $$a, then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to $$f$$ and $$g$$ on the interval $$[a,x]$$ (and a similar statement holds for $$a-\delta.

Why is the following step necessary

First applying the Mean Value Theorem to $$g$$, we see that $$g(x) \neq > 0$$, for if $$g(x)=0$$ there would be some $$x_1$$ in $$(a,x)$$ with $$g'(x_1)=0$$, contradicting $$(2)$$.

Didn't we already know this fact, precisely from $$(2)$$?

Now applying the Cauchy Mean Value Theorem to $$f$$ and $$g$$, we see that there is a number $$\alpha_x$$ in $$(a,x)$$ such that

$$[f(x)-0]g'(\alpha_x)=[g(x)-0]f'(\alpha_x)$$

or

$$\frac{f(x)}{g(x)}=\frac{f'(\alpha_x)}{g'(\alpha_x)}$$

Now $$\alpha_x$$ approaches $$a$$ as $$x$$ approaches $$a$$, because $$\alpha_x$$ is in $$(a,x)$$.

I understand this as: since $$|\alpha_x-a|<|x-a|$$ so $$0<|x-a|<\delta \implies 0<|\alpha_x-a|<\delta$$

Since we are assuming that $$\lim\limits_{y \to a} \frac{f'(y)}{g'(y)}$$ exists, it follows that

$$\lim\limits_{x \to a} \frac{f(x)}{g(x)}=\lim\limits_{x \to a} \frac{f'(\alpha_x)}{g'(\alpha_x)}=\lim\limits_{y \to a} \frac{f'(y)}{g'(y)}$$

I don't quite understand how to prove this last step.

$$\lim\limits_{x \to a} \frac{f'(\alpha_x)}{g'(\alpha_x)}=l$$ means

$$\forall \epsilon\ \exists \delta>0\ \forall x, 0<|x-a|<\delta \implies \left | \frac{f'(\alpha_x)}{g'(\alpha_x)}-l \right |<\epsilon$$

Where is the implication from $$0<|\alpha_x-a|<\delta$$ to $$\left | \frac{f'(\alpha_x)}{g'(\alpha_x)}-l \right |$$?

Didn't we already know this fact, precisely from (2)?

$$(2)$$ is about $$g'(x)$$ being non zero in a punctured interval whereas MVT application concludes that $$g(x)$$ is also non zero in that interval.

Where is the implication from $$0<|\alpha_x-a|<\delta$$ to $$\left | \frac{f'(\alpha_x)}{g'(\alpha_x)}-l \right |$$?

You agree that $$\lim\limits_{x \to a} \frac{f'(\alpha_x)}{g'(\alpha_x)}=l$$ means the following: $$\forall \epsilon>0\ \exists \delta>0\ \forall x, 0<|x-a|<\delta \implies \left | \frac{f'(\alpha_x)}{g'(\alpha_x)}-l \right |<\epsilon\tag 1$$

Since $$0<|\alpha_x-a|<|x-a|, (1)$$ also gives the following:$$\forall \epsilon>0\ \exists \delta>0\ \forall x, 0<\color{blue}{|\alpha_x-a|}<|x-a|<\delta \implies \left | \frac{f'(\alpha_x)}{g'(\alpha_x)}-l \right |<\epsilon,$$ which can be re-written as:

$$\forall \epsilon>0\ \exists \delta>0\ \forall x, 0<\color{blue}{|\alpha_x-a|}<\delta \implies \left | \frac{f'(\alpha_x)}{g'(\alpha_x)}-l \right |<\epsilon$$

which is same as saying: $$\lim_{\alpha_x\to a} \frac{f'(\alpha_x)}{g'(\alpha_x)}=l.$$ Now just substitute $$\alpha_x=y$$.

The implication comes from the definition of a limit. For each neighborhood of $$l$$, there is a neighborhood of $$a$$ where $$\frac{f'}{g'}$$ maps all elements of that neighborhood to elements in the neighborhood of $$l$$. It has been previously stated that $$a_x\in (a,x)$$, hence if $$x$$ belongs to an $$\epsilon$$ neighborhood of $$a$$, then so does $$a_x$$. Also in the last step, we don't have the implication from $$a_x$$. The implication is due to the fact that $$|x-a|<\delta$$, while $$|a_x-a|<\delta$$ is both a consequence and a useful fact, that allows us to show that $$\frac{f'(a_x)}{g'(a_x)}$$ gets close to $$l$$.