Definitions 4.32 and 4.33 and Theorem 4.34 in Baby Rudin: Are these proofs the limit statements based on these definitions correct? Here's Theorem 4.34 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $f$ and $g$ be defined on $E \subset \mathbb{R}$. Suppose $$ f(t) \to A, \ \ \ g(t) \to B, \ \mbox{ as } \ t \to x.$$ Then
(a) $f(t) \to A^\prime$ implies $A^\prime = A$.
(b) $\left( f+g \right)(t) \to A+B$.
(c) $\left( fg \right)(t) \to AB$.
(d) $\left( f/g \right)(t) \to A/B$.
provided the right members of (b), (c), and (d) are defined.
Note that $\infty - \infty$, $0 \cdot \infty$, $\infty / \infty$, $A/0$ are not defined.

Now here is Definition 4.33 in Rudin:

Let $f$ be a real function defined on $E \subset \mathbb{R}$. We say that $$ f(t) \to A \ \mbox{ as } \ t \to x,$$ where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V \cap E$ is not empty, and such that $f(t) \in U$ for all $t \in V \cap E$, $t \neq x$.

And, finally here is Definition 4.32:

For any real $c$, the set of real numbers $x$ such that $x &gt; c$ is called a neighborhood of $+\infty$ and is written $(c, +\infty)$. Similarly, the set $(-\infty, c)$ is a neighborhood of $-\infty$.

How to prove Theorem 4.34 using Definitions 4.33 and 4.32?
My effort:
Theorem 4.34(a):

If $f(t) \to A$ and $f(t) \to A^\prime$ as $t \to x$, then, for every neighborhood $U$ of $A$, we can find a neighborhood $V$ of $x$ such that $V \cap E$ is not empty, and such that $f(t) \in U$ for all $t \in V \cap E$, $t \neq x$, and for every neighborhood $U^\prime$ of $A^\prime$, we can find a neighborhood $V^\prime$ of $x$ such that $V^\prime \cap E$ is not empty, and such that $f(t) \in U^\prime$ for all $t \in V^\prime \cap E$, $t \neq x$.
Now if $A \neq A^\prime$, then we can find neighborhoods $U$ and $U^\prime$ of $A$ and $A^\prime$, respectively, such that $U \cap U^\prime$ is empty. Now $V \cap V^\prime$ is not empty and is a neighborhood of $x$. What next?

Theorem 4.34(b), (c), and (d):

If $f(t) \to A$ and $g(t) \to B$ as $t \to x$, then, for every neighborhood $U$ of $A$ and for every neighborhood $V$ of $B$, we can find neighborhoods $W_1$ and $W_2$, respectively, of $x$ such that $W_1 \cap E$ and $W_2 \cap E$ are not empty, and such that $f(t) \in U$ for all $t \in W_1 \cap E$, $t \neq x$, and such that $g(t) \in V$ for all $t \in W_2 \cap E$, $t \neq x$. What next?
4.34(a) is not necessarily right. Consider a function f defined on E={1,2,3} such that f(j)=j for all j, and x=2. Then according to 4.33, you can show both f(t) -&gt; 1 and f(t)-&gt; 3 when t -&gt; 2.
Specifically, for this function f, you can pick any neighborhood of 1, and of course you can always choose V=(0, 2.5), so V intersects E with {1,2}, and f(1)=1 is of course in the neighborhood of 1. A similar story can be told for any neighborhood of 3.
One way to fix the problem is to add an assumption in 4.33. For we to be able to say that f(t)-&gt; A when t-&gt; x, we add the assumption that x is a limit point of E. After that, (a) is not a problem. You can try (b,c,d).
4.34(a) is not necessarily right. Consider a function f defined on E={1,2,3} such that f(j)=j for all j, and x=2. Then according to 4.33, you can show both f(t) -&gt; 1 and f(t)-&gt; 3 when t -&gt; 2.
Specifically, for this function f, you can pick any neighborhood of 1, and of course you can always choose V=(0, 2.5), so V intersects E with {1,2}, and f(1)=1 is of course in the neighborhood of 1. A similar story can be told for any neighborhood of 3.
One way to fix the problem is to add an assumption in 4.33. For we to be able to say that f(t)-&gt; A when t-&gt; x, we add the assumption that x is a limit point of E. After that, (a) is not a problem. You can try (b,c,d).