Searching for the question doesn't seem to give any relevant answers on stack exchange or on google. I know that $\liminf_{n \rightarrow \infty} f(x_n)$ is always unique, but I haven't seen a proof for it. Here is my attempt:
Assume $l, l'$ are two valid values of $\liminf_{n \rightarrow \infty} f(x_n)$ for some real valued function $f$ and a sequence $x_n$ that converges to $x_0$ as $n\rightarrow \infty$. Without loss of generality, assume that $l'=l+\eta$ for some $\eta > 0$. From the definition of $\liminf$, we have that for any fixed $\epsilon> 0$, there exists a $\delta > 0$ such that for all $x \in N_\delta(x_0)$: $f(x) > l - \epsilon$ and $f(x) > l' - \epsilon$. Adding these inequalities together and substituting the value of $l'$:
$$f(x) > l - \epsilon + \eta / 2$$
Let $\epsilon = \eta / 2$ so that the above inequality is: $f(x) > l$. But this is a contradiction since the $\liminf$ is the greatest lower bound: for some $x \in N_\delta (x_0)$ we have that $f(x) < l + \eta / 2$ and $\eta$ can be made arbitrarily small.
Is this last point correct? Is the result truly a contradiction or am I missing a step?
I am giving you a demonstration , I hope it helps you.
Suppose $f(x_n)$ has at least two limits $M,L$ such that $L<M$. Considering $\epsilon = \frac{M-L}{2}$, It is well defined because $\epsilon > 0 $.As L is a Limit of $f(x_n)$ exists $N_1$ such that:
$$\forall n \geq N_1 , L - \epsilon < f(x_n) < L +\epsilon.$$
And As M is a Limit of $f(x_n)$ exists $N_2$ such that:
$$\forall n \geq N_2 , M - \epsilon < f(x_n) < M +\epsilon.$$
Defining $N = \max{(N_1,N_2)}$ therefore $f(x_N)$ verify both inequalities, in particular :
$$M - \epsilon < f(x_N) < L +\epsilon.$$
As $\epsilon = \frac{M-L}{2} \Rightarrow M-\epsilon = \frac {M+L}{2}$ and $L+\epsilon = \frac{M+L}{2}$, finally:
$$ \frac{M+L}{2}$ < f(x_N)< \frac{M+L}{2}$$.
In conclusion as it is a contradicction $f(x_n) $could only have one limit.