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Would I be right in stating that for first-countable spaces when finding the limit of a function it can equivalently be reduced to finding a limit involving sequences:
For example if looking for the limit of the function $$\lim\limits_{x \rightarrow a}f(x) = \lim\limits_{x \rightarrow a} x^{2}\sin(\frac{1}{x})$$ it is equivalent to consider finding the the limit of the sequence $$f(x_{n}) = (x_{n})^{2}\sin(\frac{1}{x_{n}})$$
for a sequence $x_{n} \rightarrow a$ (both of which will be 0).
You want to show that $\lim_{x \to a} f(x) = L$ iff for any sequence $x_n \to a$, you have $\lim_{n} f(x_n) = L$.
The $\Rightarrow$ direction is straightforward.
For the $\Leftarrow$ direction, use the contrapositive.
Then there exists some $\epsilon>0$ such that for any neighbourhood $U$ of $a$, we have some $x \in U$ such that $|f(x)-L| \ge \epsilon$.
Now construct a sequence $x_n \to a$ using a local base: Let $B_n$ be a countable base at $a$ and let $U_n = B_1 \cap \cdots \cap B_n$, which is open. If we choose any $x_n \in U_n$, then it is straightforward to see that $x_n \to a$.
Now choose $x_n \in U_n$ to be the element such that $|f(x_n)-L| \ge \epsilon$.
Hence there exists some $\epsilon>0$ and a sequence $x_n \to a$ such that such that $|f(x_n) - L| \ge \epsilon$ for all $n$.
