Where is the transition of sequences and functions?
With properties I mean for example rules like the sandwich theorem
$$\lim_{n\rightarrow\infty}a_n=K, \lim_{n\rightarrow\infty}c_n=K\text{ and }a_n\leq b_n\leq c_n\forall_{n\in\mathbb{N}}\Rightarrow\lim_{n\rightarrow\infty}b_n=K$$
Or that one can "pull out" the constant, i.e
$$\lim_{n\rightarrow\infty}a_n=K\Rightarrow \lim_{n\rightarrow\infty}ca_n=c\lim_{n\rightarrow\infty}a_n=cK$$
To name a few.
We go into some generality, when we approach what I would call the link between sequence and function limits.
Sequence limits are given by the $\epsilon-N$ definition and function limits by the $\epsilon-\delta$ definition, if you recall.
Well, first, on the real line, let us prove the following proposition (I am calling this SFR, but you can give it any name you like : it is just nice):
SFR (Sequence-Function Relation) : Let $f : \mathbb R \to \mathbb R$ and $x_0,L \in \mathbb R$. The following are equivalent :
$(1)\displaystyle\lim_{x \to x_0} f(x) = L$.
$(2)$ For every sequence $\{x_n\} \subset \mathbb R$ such that $x_n$ converges to $x_0$, we have that the sequence $\{f(x_n)\}$ converges to $L$.
Notation : $a_n \to b$ means "the sequence $a_n$ converges to $b$".
Proof : Suppose $\lim_{x \to x_0} f(x) = L$. Fix a sequence $y_n \to x_0$. We want to show that $f(y_n) \to L$. For this , we start with $\epsilon > 0$.
For that $\epsilon$, we get a $\delta > 0$ from $(1)$ such that whenever $|x_0-z| < \delta$ we have $|L-f(z)| < \epsilon$.
Now, for this $\delta > 0$ we get from the fact that $y_n \to x_0$, an $N \in \mathbb N$ such that $m > N \implies |y_n -x_0| < \delta$.
Therefore : $$ m > N \implies |y_n - x_0| < \delta \implies |f(y_n) - f(x_0)| < \epsilon $$
and hence $f(y_n) \to f(x_0)$.
For the other way, we assume that $(1)$ is not true. So $\lim_{x \to x_0} f(x) \neq L$. We now have to find a sequence $y_n \to x_0$ such that $f(y_n) \not \to L$.
For this, we "negate" the definition of the limit existing : since the limit says that for all $\epsilon > 0$ something happens, if the limit does not exist then for some $\epsilon_0 > 0$ that something does not happen.
What does not happen? "I can find a $\delta > 0$ so that ... happens" does not happen. In other words, you can't find a $\delta > 0$ for this $\epsilon_0$. Or, every $\delta > 0$ fails for this $\epsilon_0$.
And what does $\delta$ being a failure mean? It means that for some point $x$ such that $|x-x_0| < \delta$, we have $|f(x) - L|$ is not less than $\epsilon_0$, or in other words $|f(x) - L| \geq \epsilon_0$.
So, $\lim_{x \to x_0} f(x) \neq L$ means exactly this :
there exists $\epsilon_0 > 0$ such that for all $\delta > 0$, there exists $x$ such that $|x - x_0| < \delta$ but $|f(x) - L| \geq \epsilon$.
Geometrically, this means : there is an $\epsilon_0 >0$ such that you can find points as close to $x_0$ as you want, with the function value at that point at least $\epsilon_0$ away from $L$.
With these tools, in the boxed quotes we take $\delta = \frac 1n$. For each such $\delta$ we will get a point $x_n$, such that $|x_n - x_0| < \delta_n = \frac 1n$ but $|f(x_n) - L| \geq \epsilon_0$. (We don't care about the value of $\epsilon_0$ : it is positive, that is what matters).
Question : does $x_n \to x_0$? The answer is of course, since we are choosing $x_n$ so that $|x_n - x_0| < \frac 1n$, so from the squeeze theorem we get $0 \leq |x_n - x_0| < \frac 1n$ telling us that $x_n - x_0 \to 0$ or that $x_n \to x_0$.
Question : does $f(x_n) \to L$? The answer is no. The reason is, suppose it were true. Then for all $\epsilon > 0$ an $N \in \mathbb N$ should exist such that whenever $m > N$ we should have $|f(x_m) - L| < \epsilon$. But then, from the way we chose $x_m$, we know that $|f(x_m) - L| > \epsilon_0$ for all $m$. So for any $\epsilon < \epsilon_0$, we cannot find any working $\delta$, hence convergence is not possible.
This completes the proof.
With this machinery, we may "lift" sequence results to function results.
Proposition : For $c \neq 0$ , $\lim_{x \to x_0} (cf)(x) = c \times \lim_{x \to x_0} f(x)$ if the RHS exists. (For $c = 0$, the LHS exists even if the RHS does not).
Proof : We want to show the LHS exists. Call $\lim_{x \to x_0}f(x) =L$. From SFR, the LHS is equivalent to : for every sequence $x_n \to x_0$, we have to show $(cf)(x_n) \to cL$. But then , from SFR on the RHS, certainly $f(x_n) \to L$. From what we know about sequences, just multiplying by $c$ gives $(cf)(x_n) \to cL$. So the RHS exists and equals $cL$.
Squeeze Theorem : let $g(x) \leq f(x) \leq h(x)$ be functions, and $x_0,L$ be such that $\lim_{x \to x_0}g(x) = L$ and $\lim_{x \to x_0} h(x) = L$. Then, $\lim_{x \to x_0} f(x) = L$.
Proof : From SFR, we need to show that for any sequence $x_n \to x_0$ we have $f(x_n) \to L$. But then, $g(x_n) \leq f(x_n) \leq h(x_n)$ for each $n$, and by SFR on $g$ and $h$, we see that $g(x_n) \to L$ and $h(x_n) \to L$. Therefore, the squeeze theorem for sequences gives $f(x_n) \to L$, as desired.
There are many more properties that one can lift using SFR. Also, SFR applies in more general situations : I just took $\mathbb R$ for illustration, but in some more general situations, when we define what $x_n \to x_0$ and $\lim_{x \to x_0} f(x)$ mean, we will be able to provide statements like SFR , which will "lift" sequence properties to function properties nicely. Also, this lifting does not happen in all cases, but that is for another day and date.
