I was trying to find a series of differentiable functions that converge uniformly to a non differentiable function, to demonstrate the fact that uniform convergence doesn't imply anything about the differentiability of the function the sequence converges to.
I came up with this example -
$f_n(x)=\left\{ \begin{array}{c} |x| &|x|\gt\frac 1 n \\ \frac x n &|x|\leq\frac1 n \end{array} \right. $
The sequence of functions converges uniformly to $|x|$, and all $f_n$ are differentiable, but $|x|$ isn't differentiable at $x=0$.
Does this seem right?
Edit- people here helped me understand that my functions are not continuous at $x=\frac 1n$, which I forgot to check.
So a correct example will be:
$f_n(x)=\left\{ \begin{array}{c} |x| &|x|\gt\frac 1 n \\ \frac {nx^2}2+ \frac1{2n} &|x|\leq\frac1 n \end{array} \right. $
Which is continuous at $|x|=\frac1n$ and differentiable there.
Your functions are all discountinuous. I guess it is an overkill but $$f_n(x)=\sum_{k=0}^n \frac{2^k \sin(2^k \cdot x)}{3^k}$$ will do a great job, as every $f_n$ is entire, but the limit is not real differentiable anywhere.