Sequence of monotone functions problem Found a problem on "The Way of Analysis" by Strichartz.

Say $f_n$ converges to $f$ (pointwise) and each $f_n$ is increasing. 

(a) Must $f$ be increasing?
(b) What happens if each $f_n$ is strictly increasing?

To be honest I have no idea. Taking $x&lt;y$ in the domain, I'm trying to use the definition, finding appropriate $\epsilon$ to somehow deduce $f(x)\le f(y)$ but failed. Any help?
Suppose $x&lt;y$ then $f_n(x) \le f_n(y)$, in particular $f_n(y)-f_n(x) \ge 0$. Taking limits gives $f(y)-f(x) \ge 0$ (the limit of a non negative sequence must itself be non negative, which is essentially the fact that
$[0,\infty)$ is closed) and hence $f(x) \le f(y)$.

Let $f_n(x) = {1 \over n}\arctan x$. Each $f_n$ is strictly increasing, the limit $f(x)  = 0$ is not strictly increasing.

Here's another $f_n(x) = {1 \over n} {x \over 1+|x|}$.
Here's a sketch of a solution: 

Choose $x&lt;y$.  We know $f_n(x)\rightarrow f(x)$ and $f_n(y)\rightarrow f(y)$.  This means that for all $\epsilon&gt;0$ there are $N(x,\epsilon)$ and $N(y,\epsilon)$ such that if $n \geq\max\{N(x,\epsilon),N(y,\epsilon)\}$, 

$$
f(x)-\epsilon\leq f_n(x)\leq f(x)+\epsilon\\
f(y)-\epsilon\leq f_n(y)\leq f(y)+\epsilon
$$ thus using one side of each of these and the fact that $f_n(x)\leq f_n(y)$ we have

$$
f(x)-\epsilon\leq f_n(x)\leq f_n(y)\leq f(y)+\epsilon
$$ and so 

$$
f(x)\leq f(y)+2\epsilon
$$ Since this inequality holds for all $\epsilon&gt;0$, it remains true in the limit $\epsilon\rightarrow 0$. 

For the strictly increasing case, you'll end up with a statement like 

$$
f(x)&lt;f(y)+2\epsilon
$$ does this imply that the strict inequality holds in the limit $\epsilon\rightarrow 0$?