Suppose $f_n→f$ uniformly, with $f_n$ continuous for all $n$. Then $f_n(x)\to f(x_0)$ as $x\to x_0$
This is the theorem I got from notes
My key attempt to show this question is as follows:
$$\lvert f_n(x)-f(x_0)\rvert\leq |f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|<\epsilon$$
$|f_n(x)-f_n(x_0)|<\frac{\epsilon}{2}$ by continuity of $f_n$ for all $n$
$|f_n(x_0)-f(x_0)|<\frac{\epsilon}{2}$ by uniform convergence of $f_n\to f$
However, I have seen a similar theorem called the uniform limit theorem and the proof is also different by using three times the absolute value difference. Capture is below
Question 1 I figured out that what we are proving is different, and I am confused are those two theorems exactly the same theorem and have no subtle difference?
Question 2 Am I misunderstanding the final step saying. Much appreciate it if someone could write $f_n(x)\to f(x_0)$ as $x\to x_0$ in terms of some limit equation.
Much appreciate your help
I can think of two possible interpretations of "$f_n(x)\to f(x_0)$ as $x\to x_0$": $$ \lim_{x\to x_0} \lim_{n\to\infty} f_n(x) = f(x_0) \tag{1} $$ $$ \lim_{n\to\infty} \lim_{x\to x_0} f_n(x) = f(x_0) \tag{2} $$ Usually "$AAA\to BBB$ as $x\to a$" means something like $$\lim_{x\to a} CCC=BBB,$$ (where $CCC$ is related to $AAA$) so $(1)$ is the go-to choice. Other than that, note that $(2)$ is obviously true because of the continuity of $f_n$, so $(2)$ is not worth formulating as a theorem. Considering that your first line in grey is a theorem, it must mean $(1)$.
Now, back to equation $(1)$ above. Since $(f_n)$ is convergent, the inner limit equals $f(x)$, so $(1)$ is equivalent to $$ \lim_{x\to x_0} f(x) = f(x_0), $$ which is equivalent to the continuity of $f$ at $x_0$, i.e. the quoted Theorem 8.2.2.