A few intro questions about limits I'm a first year university student and it is my first time posting on the forum so if I have posted incorrectly please let me know and I'll keep it in mind for next time! Although I've already finished my first calculus class, the course moved quite quickly and so I've been revising to clear up a few points I think could use some strengthening.
While going over some $\delta-\varepsilon$ limit exercises, I realized there are few things about the $\delta-\varepsilon$ definition that either I don't understand, or would just like some confirmation about.
Firstly, why is the definition not an equivalence statement? For any $\varepsilon$ value considered wouldn't there be a corresponding $\delta$ range for $x$?
Also why is statement $0 < |x-a|< \delta$ , with $<$ rather than $\le$? With the $0<|x-a|$, is this simply to increase the strength of the statement by not requiring that the definition hold for the limit at $x=a$? But why then have $|x-a|< \delta$? Is there some issue with allowing $|x-a| \le \delta$?
I realized this was a point I really didn't understand while reviewing the definition we were given for limits as $x \to \infty$. Here we were told that this has the limit $l$ when there exists an $N$ and $\varepsilon$ such that: $x \ge N \implies |f(x)-l|< \varepsilon$. So here the statement allows for a $\le$, while the finite statement doesn't, which seemed a little curious.
 A: A clarification of some of your points of confusion:

First off, the second number us usually $\delta$, which is the letter "delta", not "sigma" ($\sigma$). Of course, you can call it whatever you like since math does not care about names.

The definition of a limit is an equivalence statement:

the function $f(x)$ has a limit of $l$ as $x$ approaches $a$ if and only if the statement $$\forall \epsilon\exists \delta\forall x: 0<|x-a|<\delta \implies |f(x) - l|<\epsilon$$

The internal statement, $0<|x-a| < \delta\implies |f(x) - f(a)|$, however, is not an equivalence, because we do not want it to be. For example, do you think the function $f(x) = x^2$ has a limit when $x$ approaches $1$? 
Well, if the statement would be an equivalence, so $|x-a|<\delta \iff |f(x) - f(a)|<\epsilon$, then $f$ does not have a limit, because you can take $a=-1$ and $x=1$ and get that $|f(x) - f(a)| = 0$, but $|x-a|=2$ which is not small! 

The statement is $0<|x-a|<\delta$, but it is equivalent to demand that $0<|x-a|\leq \delta$. This is because you can freely choose $\delta$, so if you just decrease $\delta$ a little more, the inequality $<$ will turn into $\leq$.
Similarly, the statement is $0<|f(x)-f(a)| < \epsilon$, not $\leq\epsilon$, because the statement must be true for every $\epsilon$, so for example if it is true for $\epsilon$, it is also true for $\frac\epsilon2$. The requirement $<$ is not stronger (and also not weaker) than $\leq$, we just decide to use one because it's easier to write.

The definition of the limit as $x\to\infty$ is the other way around, so it is:

For all $\epsilon > 0$ there exists some $N$ such that if $x>N$, then $|f(x) - l| < \epsilon$.

This can be understood similarly to the standard limit:


*

*The standard limit is: no matter how close I want to be to $f(a)$, if I set $x$ close enough to $a$, I know that $f(x)$ will be close to $f(a)$.

*The infinite limit: No matter how close I want to be to $f(a)$, if I set $x$ to be large enough (conceptually "if I set $x$ close to infinity"), then $f(x)$ will be close to $f(a)$.

A: Be careful: the definition of $\lim_{x \to +\infty} f(x)= \ell$ is

For every $\epsilon>0$ there exists a number $N>0$ such that $x > N$ implies $|f(x)-\ell|<\epsilon$.

What you wrote is rather different, since you require the existence of some $\epsilon$ and $N$. Also in this case, the condition $x \geq N$ is by no means different than $x>N$.
A: In general, a delta-epsilon proof is saying that for every value $x_i$ that is arbitrarily close to $X$, a function $f(x)$ will yield some value $y_i$ that is arbitrarily close to $Y$. 
The statement is: 
$$\forall \epsilon > 0, \exists\delta>0, s.t. 0<|x-x_0|<\delta \implies |f(x)-f(x_o)|< \epsilon$$
So, consider what happens in the case $x=x_o$ it is clear then that you have: $$|f(x)-f(x)|=0 < \epsilon , \forall \epsilon > 0$$ 
such a case is trivial and is often untreated (such was my experience).The intuitive interpretation should be something like this: 
If I put the same input through a function twice it will yield the same output twice. And the euclidean distance between any number and itself is 0 and therefore obviously smaller than any positive epsilon. And since my input was the same number twice, again considering that the euclidean distance between a number and itself is zero, I will have that my two inputs are within $\delta$ of each other for every $\delta>0$.  
The reason we use < instead of $\leq$ is because if the statement is true it must be true for any possible $\epsilon$ so if some $\epsilon$ qualifies because of the equality in $\leq$ then we have obviously that $\frac{\epsilon}{2}$ will fail. Since we say this must be true for all $\epsilon$, we can say the statement is not true if there is just one instance where it is untrue. 
