Taking limits on each term in inequality invalid? So this inequality came up in a proof I was going through.
$$c - 1/n < f(x_n) \leq c$$
Where $c$ is a real number, $f(x_n)$ is the image sequence of some arbitrary sequence being passed through a function and $n$ is a natural number. At this point the author simply concludes that this implies the function sequence converges to $c$. It's pretty clear that this is happening, but I'm not exactly sure what the proper justification is.
Are we taking limits as $n \to \infty$ on all sides of the inequality? So that we get 
$$c < \lim_{n \to \infty} f(x_n) \leq c$$
Can we just take limits on all sides of an inequality like that? It seems like that could lead to problems as you could take a situation like 
$$1 - 1/n < 1$$ but then doing what I suggested would just lead to $1<1$ which is not true. 
So what is the argument that I seem to be overlooking?
Thanks.

Here is another proof that completely avoids squeeze theorem
we will prove that $(f(x_n)) \rightarrow c$


*

*consider arbitrary $\epsilon>0$

*we have to show that there exists $N$ s.t forall $n\geq N$ $|f(x_n) -c | < \epsilon$

*we have that $c-\frac{1}{n} < f(x_n) \leq c \implies  -\frac{1}{n} < f(x_n)-c \leq 0$

*Now  using the fact that if $a < x \leq b$ implies $|x| \leq \max\{|a|, |b|\}$ we can say that $|f(x_n)-c| \leq \max\{\frac{1}{n},0\} \leq \frac{1}{n}$

*since $(\frac{1}{n}) \rightarrow 0$ there exists $N$ s.t forall $n\geq N$ $\,\,$ $\frac{1}{n} < \epsilon$

*we will show this $N$ indeed suffices

*consider arbitrary $n \geq N$

*$|f(x_n) -c | \leq \frac{1}{n} < \epsilon$ (last inequality comes from fact that $n \geq N$)

*Hence $\lim f(x_n) = c$
