If $\overline\lim_{x\to y} f(x)=A$ and $\langle x_n\rangle$ is a sequence with $x_n \neq y$ such that $y=\lim x_n$, then $\overline\lim_{n\to \infty} f(x_n)\le A$.
Now, I am able to visualize this problem graphically. One instance where
$\overline\lim_{n\to \infty} f(x_n)= A$ is when $f$ is continuous over a certain neighborhood of $y$, since the graph is "smooth" around that point, all $f(x)$'s of different sequences converge to the same $f(y)$. But if $f$ is not continuous, the limit may reach different values depending on the "direction" $x$ converges to $y$. (Hopefully this made sense.)
What I don't know how to do is write out the proof mathematically and simply. Now, my textbook defined the limit superior as follows:
$$\overline\lim_{x\to y}f(x)=\inf_{\delta>0}\sup_{0<|x-y|<\delta} f(x)$$ $$\overline\lim_{n\to \infty}x_n= \inf_{n}\sup_{k \ge n} x_k$$
While I understand what these mean, I'm having trouble to express them in mathematical terms, and even if I can, I'm not sure how to proceed. Can someone explain to me how to express these things correctly and solve the problem in a good way? Thanks.
Let $\epsilon > 0$. Since $\varlimsup\limits_{x\to y} f(x) = A$, there exists a $\delta > 0$ such that $\sup\limits_{0 < \lvert x - y\rvert < \delta} f(x) < A + \epsilon$. As $x_n \to y$ and $x_n \neq y$ for all $n$, there exists a positive integer $N$ such that for all $n \ge N$, $0 <\lvert x_n - y\rvert < \delta$. Hence, for all $n \ge N$, $$f(x_n) \le \sup_{0 < \lvert x - y\rvert < \delta}f(x) < A + \epsilon$$ Therefore $$\varlimsup_{n\to \infty} f(x_n) \le \sup_{n\ge N} f(x_n) \le A + \epsilon$$ Since $\epsilon$ was arbitrary, $\varlimsup\limits_{n\to \infty} f(x_n) \le A$.