When does $ f'_{n}(x) \to g(x) =1$ imply $f'(x) =1 $ I considered the following:
$f_{n}(x) \in C^1(0,1)$ (class of continuously differentiable functions) and $f_{n} \to f:(0,1) \to \bf{R} $ with $f'_{n} \to g =1$. Does this imply that $f \in C^1(0,1)$ also? The convergence is pointwise.
I realized that the function sequence $f_{n}(x) = \arctan(nx)+x$ is a counterexample. However, suppose that the convergence is in $L^1$, i.e., in the mean $p$ with $p=1$. Can we now claim that $f \in C^1(0,1)$ and $f'(x) =1 $?
 A: Yes, convergence $f_n'\to 1$ in $L^1$ is enough to conclude that $f(x)=x$ up to a constant. Indeed, let $c = f(1/2) = \lim_{n\to\infty} f_n(1/2)$. The fundamental theorem of calculus implies
$$
|f_n(x) - (x-1/2+c)| \le |f_n(1/2) - c| + \|f'_n-1\|_{L^1} \to 0
$$
Thus, $f_n(x) \to x-1/2+c$ uniformly. The function $f(x) = x-1/2+c$ is obviously in $C^1$. 
More generally (if $g$ wasn't identically $1$) under the $L^1$ convergence assumption you can conclude that $f$ is an antiderivative of $g$. So, as long as $g$ is continuous, $f\in C^1$ follows. 

Aside. Because I did not really understand your example with $\arctan$, I wrote up another one. With two minor modifications in the setup: 


*

*Let's subtract $x$ from every $f_n$, so that we have simpler condition $f_n'\to 0$ instead of $f_n'\to 1$. 

*Let's replace $(0,1)$ with $(-1,1)$ so that it's easier to create bad things insiden .


Define $f_n(x) = e^{-nx^2}$. As $n\to \infty$, this sequence converges pointwise to the discontinuous function 
$$f(x) = \begin{cases} 1 \quad & x=0, \\ 0 \quad & x\ne 0 \end{cases} $$
Yet, the sequence of derivatives, $f'_n(x) = -2xn e^{-nx^2}$ converge to $0$ pointwise. 
