Definition of $\lim_{x \to x_0} f_n (x)$ and related questions

Question

While studying uniform convergence of sequences of functions, I stumbled upon the theorem that, under uniform convergence hypothesis of $f_n$ to $f$ as $n \to \infty$, one can interchange the limits: $$\lim_{n \to \infty} \lim_{x \to x_0} f_n (x)=\lim_{x \to x_0} \lim_{n \to \infty} f_n (x).$$ However, on my textbook there is no definition of what $\lim_{x \to x_0} f_n (x)$ means. So I tried to formulate a definition, trying to adapt it from multivariable calculus (since the notaion $a_n$ is just another way to express the function $a:\mathbb{N} \to \mathbb{R}$ defined by $a_n=a(n)$). In the end, since for $x \in \mathbb{R}$ the sequence of functions $f_n(x)=f(n,x)$ can be seen as a function $f:\mathbb{N} \times \mathbb{R} \to \mathbb{R}$, I tried a more general case: for $X,Y \subseteq \mathbb{R}$, it is $\lim_{x \to x_0} f(x,y)=l(y)$ if for any $\epsilon>0$ and for any $y \in Y$ there exists $\delta_{\epsilon,y}>0$ such that for any $x \in (X\cap(x_0-\delta_{\epsilon,y},x_0+\delta_{\epsilon,y)} \setminus \{x_0\})\times Y$, $|x-x_0|<\delta_{\epsilon,y} \implies |f(x,y)-l(y)|<\epsilon$.

Question 1: Is this definition the correct one that is accepted for what is means for $\lim_{x \to x_0} f(x,y)=l(y)$? I am not sure, in particular, about the quantification of $y$; I believe this is ok because quantifying $y$ after $\delta$ would give me osme kind of uniform limit in $y$ and this is not I want to define, and of course I want this to hold no matter what $y$ I take and so quantify $y$ before $\delta$ and with "for any $y$" seemed to me the only reasonable possibility.

If the definition given above is correct, another question that I was thinking of is the following: consider $f_n(x)=nx$. If I want to evaluate $\lim_{x \to 0} nx$ I can say that, given $\epsilon>0$, if $\delta_{\epsilon,n}=\frac{1}{n}\epsilon$ then $|x|<\delta_{\epsilon,n}=\frac{1}{n}\epsilon \implies |nx| < \epsilon$ and so $\lim_{x \to 0} nx=0$. However, since $n$ is independent on $x$, maybe I can write $\lim_{x \to x_0} nx =n \cdot \lim_{x \to 0} x$ and the latter limit is $0$ with if $\delta_{\epsilon}=\epsilon$, which is independent on $n$, differently from before.

Question 2: What is happening in this latter limit about the dependence of $\delta$ in the two different approaches? I tend to conclude that one is wrong, or the definition I came up with is wrong. Another reason I thought about is that the equality $\lim_{x \to 0} nx=n \cdot \lim_{x \to 0} x$ may bot hold, in particular if I fix $n=1$ and $n=2$ I need in the first case $\delta<\epsilon$ and in the second case $\delta<\frac{1}{2}\epsilon$ and so it seems like that $n$ has a role in the choice of $\delta$ that can't be ignored. Maybe this doesn't hold because, even if $|x|<\epsilon$ for arbitrary $\epsilon>0$, when multiplied by $n$ that varies in the unbounded set $\mathbb{N}$ the product $nx$ could not be arbitrary snall and so taking $n$ outside the limit doesn't work for this reason? Or is there any reason why?

Answer 1

Question 1 Your definition is correct (except that it should be $x\in X\cap (x_0 - \delta_{\epsilon,y}, x_0 + \delta_{\epsilon,y})$, without the $\times Y$). An equivalent way of stating this is that for any $y\in Y$, the partial function $f_y : x\in X \mapsto f(x,y)$ has $\lim_{x\to x_0} f_y(x) = l(y)$, where the limit is the usual one from calculus.

Using this definition, we see that your calculation is valid : $$\lim_{x\to 0} nx = n\lim_{x\to 0} x = n\times 0 = 0 $$

Question 2 When trying to evaluate an expression like : $$\ell = \lim_{n\to +\infty} \lim_{x\to 0} (nx)$$ you need to first compute the rightmost limit: define a sequence $(a_n)$ by : $$ \forall n\in \mathbb N, a_n = \lim_{x\to 0} nx \tag 1$$ and then : $$\ell = \lim_{n\to +\infty} a_n\tag 2$$

When computing $a_n$ in $(1)$, $n$ is fixed and arbitrary. The limit is then defined as : $$\forall n\in \mathbb N,\forall \epsilon >0,\exists \delta >0, \forall x\in(-\delta,\delta) , |nx|<\epsilon \tag{1'}$$ The quantifier for $n$ is in front, so $\delta$ can be dependent on it. It doesn't matter.

Then, once we proved $(1')$, we can go on to prove $(2)$ which is defined as : $$\forall \epsilon >0, \exists N\in\mathbb N, \forall n\geq N, |a_n-\ell|<\epsilon\tag{2'} $$

The fact that we cannot eliminate the dependence of $\delta$ on $n$ in $(1')$ shows that the convergence is not uniform. It does matter when we only consider point-wise convergence.

Answer 2

You are greatly over-thinking this. The indices $n$ are just labels. You have a sequence of functions $\{f_1, f_2, f_3,\dots\}$. For example, you might have $f_1(x)=x$, $f_2(x)=x^2$, and $f_n(x)=x^n$. For any of the $f_n$, that is, with $n$ fixed, the sentence $\lim_{x\to x_0}f_n(x)=\lim_{x\to x_0}x^n=L$ makes perfect sense without thinking of the sequence of functions as a function on $X\times\mathbb{N}$.

As to your second question, about $\delta_\epsilon=\epsilon$ being independent of $n$, this is true but not really relevant. The sequence $f_n(x)=nx$ indeed converges pointwise to zero, and your proof of this is correct. However, I just wanted to be sure that you understand that this is not, on any interval, an example of uniform convergence of the sequence $\{f_n\}$ to zero.