What is really meant by "letting a variable tends to" 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 $\delta&gt;0$ 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 $x-\delta&lt;t&lt;x$ 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.
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 $x+\delta&lt;t&lt;x$ is impossible if $\delta&gt;0$. I expect they mean $t$ is larger than $x$ and we let $t\to x$ from above.
For $t&lt;x$ 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 $f'(x) \ge 0$.