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.

Its a pretty standard theorem which was originally called "Sandwich Theorem" but nowadays more popularly known as "Squeeze Theorem".
Squeeze Theorem: If $\{a_{n}\}, \{b_{n}\}, \{c_{n}\}$ are sequences such that $a_{n} \leq b_{n} \leq c_{n}$ for all $n$ greater than some specific positive integer $N$ and $\lim_{n \to \infty}a_{n} = \lim_{n \to \infty}c_{n} = L$ then the sequence $b_{n}$ also converges to $L$. The conclusion is same even if one or both of the inequalities in $a_{n} \leq b_{n} \leq c_{n}$ are replaced by their strict versions.
In the current question $a_{n} = c - (1/n), b_{n} = f(x_{n}), c_{n} = c$ so that $c_{n}$ is a constant sequence. Clearly both the sequences $a_{n}, c_{n}$ converge to $c$ and hence $b_{n} = f(x_{n})$ also converges to $c$.
The proof of the Squeeze theorem is particularly easy. Since both the sequences $a_{n}, c_{n}$ tend to $L$ as $n \to \infty$ it is possible to find a positive integer $m > N$ corresponding to any given $\epsilon > 0$ such that $$L - \epsilon < a_{n} < L + \epsilon,\, L - \epsilon < c_{n} < L + \epsilon$$ for all $n \geq m$. Since the value $m$ is chosen to be greater than $N$, the inequalities between $a_{n}, b_{n}, c_{n}$ hold for $n \geq m$ and hence we have $$L - \epsilon < a_{n} \leq b_{n} \leq c_{n} < L + \epsilon$$ for all $n \geq m$. And this means that $$L - \epsilon < b_{n} < L + \epsilon$$ for all $n \geq m$ (you can clearly see that this conclusion remains same even if we had strict inequalities between the given sequences). Therefore $b_{n} \to L$ as $n \to \infty$.

Your problem (or confusion) is also related to the application of limit operation on inequalities.
If we have sequences $a_{n}, b_{n}$ with $a_{n} < b_{n}$ and both the sequences converge to $a, b$ respectively then from this information we can only conclude that $a \leq b$.
Since $a \leq b$ includes both the cases $a = b$ and $a < b$, both the options are possible in actual examples. For example take $a_{n} = 1 - (1/n), b_{n} = 1 + (1/n)$ then we have $a_{n} < b_{n}$ and both sequences tends to $1$. On the other hand if $a_{n} = 1/n$ and $b_{n} = 1 + (1/n)$ then we have the case $a = 0, b = 1$ so that $a < b$. Normally we need more information to find whether $a < b$ or $a = b$.
