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This is the entire proof for the 0/0 case and I am very lost. I understand choosing $r>A$, but I don't understand why we're choosing $q$. I can see why 18 is true, but I do not understand how the strict inequality becomes a weak inequality going from (18) to (19). Finally, and perhaps most importantly, I don't understand how (19) shows that $f(x)/g(x)\rightarrow A$ as $x\rightarrow a$.

Can someone help me follow this through?

  • 2
    Here's a [better proof](http://math.stackexchange.com/questions/1798131/proof-of-lhopitals-rule/1993679#1993679). This Rudin fellow makes me want to vomit.2017-03-18
  • 0
    For the last question: it is not that (19) alone shows the limit. One should combine (19) and (22) together.2017-08-21

2 Answers 2

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That's not the entire proof. It's half of it; the other half is on the next page (although that's definitely not clear from the way it's written).

What Rudin is doing is choosing two points $p,q$ with $p

The use of $r$ (and other similar choices in the proof) is usually due to the fact that the theorems he is quoting require open intervals.

The inequality thing from (18) to (19) is the well-known fact that if you have $x_n

  • 0
    The next page deals with the $\infty / \infty$ case. I'll add it though.2012-12-04
  • 1
    I know and, as I said, Rudin's proof is not very well written. Unless you are prepared to admit that $f(y)/g(y)\leq A$ implies that $f(y)/g(y)\to A$.2012-12-04
  • 1
    Okay, after staring at this for about 24 hours I have finally convinced myself that it makes sense. Thanks very much for your help!2012-12-05
  • 1
    You are welcome. I recommend you try reading the proof from another source; it is not a nice proof to write, but in my view Rudin's writing is particularly cumbersome.2012-12-05
  • 0
    Do you have any recommended sources?2012-12-08
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I actually quite like this proof after puzzling through it for a while. I'll only address the first half of the proof, and I can answer each of your questions about

  1. Why we are choosing $r$ and $q$ and
  2. In going from (18) to (19), how we go from strict to non-strict inequality.

First to lay some groundwork. (I'm going to quote Rudin's Definition 4.33 first.) Let $f$ be a real function defined on $E\subset\mathbf R$. We say that $$ f(t)\to A\:\text{as}\:t\to x, $$ where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V\cap E$ is not empty, and such that $f(t)\in U$, for all $t\in V\cap E$, $t\ne x$.

In the first case, we are considering the case where $A$ could be $-\infty$ or some real number. The idea is we are trying to show that for every neighborhood $U$ of $A$ there is some neighborhood $(a,c)$ of $a$ such that we can squeeze the quotient $f(y)/g(y)$ past the right endpoint of $U$ into $U$. Think of the neighborhood $U$ as having right endpoint $q$.

Thus, we choose $A

To address 2., suppose we have a framework as in Rudin's Definition 4.33 and furthermore that $f(x)M$, then the idea is we should be able to find some neighborhood around $\lim f(x)$ such that in that neighborhood all the values $f(x)$ are greater than $M$, a clear contradiction of our hypothesis that $f(x)< M$ on $E$. From Rudin's (18), define $$ h(x) = \frac{f(x)-f(y)}{g(x)-g(y)}. $$ If you agree that $h(x)neighborhoods of $A$, but our non-strict inequality is preventing that. This is where $q$ comes in. If we note that for arbitrary $q>r$, we could get $f(y)/g(y)