I have a certain need of rigorously justify some steps of proofs that use limits (maybe because I don't want to rely on intuition about limits that I have from first courses of calculus or somewhat), but I'm not sure what to think when I read "let $x\to a$".

I not sure if I'll make myself clear. I have done basic courses of calculus and I'm actually studying undergraduate level Real Analysis. I know basic properties of limits and know how to prove limits using epsilons and deltas. I'm interested in whether I can use the term "tending to" as it is used, for example, in the following proof of Fermat's Theorem (this proof is given on theorem 5.8 of Rudin's Principles of Mathematical Analysis):

Theorem: Let $f$ be defined on $[a,b]$; if $f$ has a local maximum at a point $x\in(a,b)$, and if $f'(x)$ exists, then $f'(x)=0$.

Beginning of the proof: Choose $\delta>0$ so that $f(x)$ is a local maximum, that is, $f(x)\ge f(t)$ for all $t\in (x-\delta, x+\delta)$. Then take $x-\delta<t<x$ then $$\frac{f(t)-f(x)}{t-x}\ge 0.$$ Letting $t\to x$, we have $f'(x)\ge 0$.

I do know I can assume that $f'(x)$ is negative and choose $\epsilon = -f'(x)$ so that the quotient will also be negative, getting a contradiction hence we must have $f'(x)\ge 0.$ But this is quite non-practical as the term "tending to" suggests. Any answer that helps me "translate" the term into rigorous thought would be appreciated.

  • $\begingroup$ In this context it means "applying $\lim_{t \rightarrow x}$ to both sides." $\endgroup$ – goblin Apr 9 '18 at 2:18
  • $\begingroup$ He's appling the limit as $t \to x$ on both sides and using the comparison theorem. It is frequently referred to as "taking the limit of an inequality" $\endgroup$ – David Reed Apr 9 '18 at 2:22

To clarify some things, I'm not exactly sure what it means to choose a negative epsilon considering the traditional convention: $|f(x_2)-f(x_1)|<\epsilon$, and, for my own bias, I tend to view $\delta$ as $\delta(\epsilon)$, being, delta as a function of epsilon as the traditional continuity definition takes $|f(x_2)-f(x_1)|<\epsilon$ for any positive epsilon, so as epsilon decreases, delta follows. Notice how we sometimes establish delta in terms of epsilon. Of course, there's nothing wrong with establishing epsilon in terms of delta either, it's more or less the same thing.

Now, I believe I could offer some intuition on what it means to "take the limit", but I'm not sure exactly what you're asking for.

Taking the limit implies continuity, for example, let's take a look at Cauchy's convergence criterion, the criterion is this:

For any two elements belonging to the same sequence $a_n$ and $a_m$, convergence is defined such that $|a_m-a_n|<\epsilon(N)$ for any positive $\epsilon$, $\epsilon$ as a function of $N$ such that $m>N$ and $n>N$. Notice the magnificence of taking epsilon as a function of $N$, that is, have $N$ as far in into the sequence as you want (have $\epsilon$ as small as you want), Cauchy's convergence criterion states that the criterion $|a_m-a_n|<\epsilon(N)$ for any positive $\epsilon$ will always be satisfied - we can take the limit with implications that the limit point can always be approximated with a greater and greater degree of accuracy (or, a lesser and lesser degree of tolerance) - that is, of course, what a limit of a subsequence is.

Incidentally, this demonstrates continuity at the limit point.

  • $\begingroup$ I actually meant $\epsilon = -f'(x)$. $\endgroup$ – AnalyticHarmony Apr 9 '18 at 2:51
  • $\begingroup$ @AnalyticHarmony I'm sorry, I don't know what that means. $\endgroup$ – user521846 Apr 9 '18 at 3:44

Essentially, it means taking the limit of $t\to x$. So if we have

$$\frac{f(t)-f(x)}{t-x}\ge 0$$

and we let $t\to x$, then we mean

$$\lim_{t\to x}\frac{f(t)-f(x)}{t-x}\ge 0$$

or $f'(0)\geq 0$.

I also suspect you've made a typo there since $x+\delta<t<x$ is impossible if $\delta>0$. I expect they mean $t$ is larger than $x$ and we let $t\to x$ from above.

  • $\begingroup$ And what is meant by "taking the limit"? $\endgroup$ – AnalyticHarmony Mar 22 '18 at 9:05
  • $\begingroup$ It means $\lim_{t \to x}\frac{f(t)-f(x)}{t-x}$. $\endgroup$ – Fred Mar 22 '18 at 9:05
  • $\begingroup$ What @Fred says. If you wish to be very rigorous and don't want to use limits, you should take a look at the definition of limits. $\endgroup$ – vrugtehagel Mar 22 '18 at 9:07

For $t<x$ we have $\frac{f(t)-f(x)}{t-x}\ge 0.$ Since $f'(x)= \lim_{t \to x}\frac{f(t)-f(x)}{t-x}$, it follows that $f'(x) \ge 0$.

  • $\begingroup$ It says $f(x)\ge f(t)$ for all $t\in (x-\delta, x+\delta)$ though so $\frac{f(t)-f(x)}{t-x}\ge 0$ is always true in that interval, not only when $t<x$. $\endgroup$ – vrugtehagel Mar 22 '18 at 9:06

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