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.
 A: Here's what the author is using:
Suppose $\{x_n\}$ and $\{y_n\}$ are sequences that converge to $x$ and $y$ respectively and for all $n$, $x_n \leq y_n$. Then $x \leq y$.
The way we get around the example in your question is to avoid the strict inequality. Things do go awry when use $<$. However, with $\leq$, things do not.
A proof of the statement, for any non-believers: it suffices to show that this is true when $x_n = 0$ for all $n$. (Just subtract the $x_n$ over if this is not the case). Suppose that $y < 0$. Then there is some sufficiently large $N$ so that if $n > N$, $|y_n -y| < -y/2$. Let $n>N$. But this would imply that $y/2< y_n - y < -y/2$. Furthermore, it would imply that
$$
y_n < y - \frac{y}{2} = \frac{y}{2} < 0.
$$
But we assumed that $0 \leq y_n$ for all $n$, so our assumption that $y < 0 $ must be false, and in fact $0 \leq y$.
A: This is an application of a basic theorem called squeeze theorem:
$\def\nn{\mathbb{N}}$

Given any sequences $(a_n)_{n\in\nn}$ and $(b_n)_{n\in\nn}$ and $(c_n)_{n\in\nn}$ such that $a_n \le b_n \le c_n$ for any $n\in\nn$:
  If $\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = x$ for some $x$:
    $\lim_{n\to\infty} b_n = x$.

Incidentally, the usual proof of the squeeze theorem (which others have already given) essentially proves a more general theorem:

Given any sequences $(a_n)_{n\in\nn}$ and $(b_n)_{n\in\nn}$ such that $a_n \le b_n$ for any $n\in\nn$:
  $\limsup_{n\to\infty} a_n \le \liminf_{n\to\infty} b_n$.

This theorem is very useful because it implies the following powerful technique in proving limits:

Given any sequences $(a_n)_{n\in\nn}$ and sequences $(b_{m,n})_{m,n\in\nn}$:
  If for any $m \in \nn$ we have $| a_n - b_{m,n} | \le \frac{1}{m}$ for any $n\in\nn$:
    $\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_{n,n}$.

A: Basically what happens is that for every $n$, $f(x_n)>c-1/n$. Maybe you're not understanding what taking a limit means. Taking the limit when $n$ approaches infinity means that:
For every $\epsilon>0$ there is an $N>0$ such that, for every $n>N$, $\left | f(x_n)-c \right |<\epsilon$.
Now, take any $\epsilon$ you want. Because $c-1/n<f(x_n)\leq c$, we can choose $N=1/\epsilon$, so that $\left |f(x_n)-c \right | < (c-\epsilon)-c= \epsilon $. We've proved, then, that $\lim_{x\rightarrow \infty }f(x_n)=c$
Now, you cannot take limits at both sides of a strict inequality like you tried to do. That would result in weird results like the one you mentioned. 
A: 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 $a = b$.
A: 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 $\lim f(x_n) = c$
