Consider this example:
$$
3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots
$$
It alternates between something approaching $3$ from below and something approaching $5$ from above. The lim inf is $3$ and the lim sup is $5$.
The inf of the whole sequence is $3-\frac12$.
If you throw away the first term or the first two terms, the inf of what's left is $3-\frac14$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac16$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac18$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac1{10}$.
. . . and so on. You see that these infs are getting bigger.
If you look at the sequence of infs, their sup is $3$.
Thus the lim inf is the sup of the sequence of infs of all tail-ends of the sequence. In mathematical notation,
$$
\begin{align}
\liminf_{n\to\infty} a_n & = \sup_{n=1,2,3,\ldots} \inf_{m=n,n+1,n+2,\ldots} a_m \\[12pt]
& = \sup_{n=1,2,3,\ldots} \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} : n=1,2,3,\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\{ a_m : m\ge n\} : n=1,2,3,\ldots \right\}.
\end{align}
$$
Just as the lim inf is a sup of infs, so the lim sup is an inf of sups.
One can also say that $L=\liminf\limits_{n\to\infty} a_n$ precisely if for all $\varepsilon>0$, no matter how small, there exists an index $N$ so large that for all $n\ge N$, $a_n>L-\varepsilon$, and $L$ is the largest number for which this holds.
What you're trying to prove, if you want to use the definition of a limit directly (at least in the $+\infty$ case) is that
For any $M>0$: there is an $N$ such that for any $n > N$, $s_n > M$
From your knowledge of the liminf, you know that
For any $M >0$: there is an $N$ such that $\inf \{s_n:n>N\} > M$
Why does this second statement imply the first?
For $\pm \infty$, you will need only the liminf or the limsup. For finite limits, however, you will need both.
The conclusion we want to reach is
For any $\epsilon > 0$: there is an $N$ such that $n>N$ implies $|s_n - L| < \epsilon$.
Using the liminf and limsup, we have the statements:
For any $\epsilon > 0$: there is an $N$ such that $n > N$ implies $\inf \{s_n : n>N\} > L - \epsilon$
For any $\epsilon > 0$: there is an $N$ such that $n > N$ implies $\sup \{s_n : n > N\} < L + \epsilon$
How do these two statements let you deduce the first?
You could also use the squeeze theorem if you note that
$$
\inf \{s_n: n \geq N\} \leq s_N \leq \sup\{s_n: n \geq N\}
$$
and take the limit as $N \to \infty$.
Best Answer
For each $n=1,2,...$, choose $y_{n}$ such that $d(x,y_{n})<1/n$ and $\inf_{y:d(x,y)<1/n}f(y)\leq f(y_{n})\leq\inf_{y:d(x,y)<1/n}f(y)+1/n$. Note that $(\inf_{y:d(x,y)<1/n}f(y))_{n=1}^{\infty}$ is monotone, then $\lim_{n}f(y_{n})=\lim_{n}\inf_{y:d(x,y)<1/n}f(y)$. Note that we also have $f(y_{n})\leq\inf_{y:d(x,y)<1/n}f(y)+1/n\leq\sup_{\delta>0}\inf_{y:d(x,y)<\delta}f(y)+1/n$ and hence $\sup_{\delta>0}\inf_{y:d(x,y)<\delta}f(y)\geq\lim_{n}f(y_{n})=\liminf_{n}f(y_{n})\geq\inf\{\liminf_{n}f(x_{n}): x_{n}\rightarrow x\}$.
On the other hand, for any $x_{n}\rightarrow x$ and $\delta>0$, then some $n_{0}$ is such that $d(x_{n},x)<\delta$ for $n\geq n_{0}$, then $f(x_{n})\geq\inf_{y:d(x,y)<\delta}f(y)$ for all such $n$ and hence $\liminf_{n}f(x_{n})\geq\inf_{y:d(x,y)<\delta}f(y)$. Since this is true for all $\delta>0$, we have $\liminf_{n}f(x_{n})\geq\sup_{\delta>0}\inf_{y:d(x,y)<\delta}f(y)$, and finally $\inf\{\liminf_{n}f(x_{n}): x_{n}\rightarrow x\}\geq\sup_{\delta>0}\inf_{y:d(x,y)<\delta}f(y).$