A couple basic questions about pointwise convergence I am looking for a few counterexamples regarding sequences of functions that converge pointwise.
Firstly, is there a sequence $f_{n}$ where all functions are differentiable and whose limit $f$ is continuous but not differentiable. The example I could think of that broke differentiability also broke continuity.
Second, is there a sequence where $f_n$ is Riemann integrable for all $n$. but $f$ is not Riemann integrable?
 A: For the first part, consider for example
$$ f_n(x) = \frac1n\sqrt{(nx)^2+1} $$
For the second, how about
$$ f_n(x) = \begin{cases} 1 & \text{if }x=p/q\text{ with }q<n \\ 0 & \text{otherwise}  \end{cases} $$
A: For the second part choose $f_n(x)=x^{\frac{1}{n}-1}$ on $(0,1]$.
A: Henning Makholm gave an example where the derivative was discontinuous in $0$, here is an example where the derivative goes to infinity on the bounds of the domain.
$f_n=\sqrt{1+\frac 1n-x^2}$ defined on $[-1,1]$.
This is a semi-circle of radius slightly bigger than $1$ so on $[-1,1]$ the function $f_n$ is perfectly differentiable.
But the limit (the semi-circle) of radius $1$ has vertical semi-tangents in $\pm 1$.
For the Riemann integrability why not take a continuous function that has a Dirac as a limit?
$f_n:\begin{cases}\dfrac{(2n+1)!}{2^{2n+1}(n!)^2}(1-x^2)^n & x\in[-1,1]\\0 & \text{otherwise}\end{cases}$ 
With $\displaystyle \int_{-1}^1f_n(t)\mathop{dt}=1$ but $f$ is not bounded so not Riemann integrable.
I admit though that $f$ not being a proper function, may not fit well the example of pointwise convergence. Nightgap's example falls in the same area. 
The indicator of rationnals given by H.M is bounded, but the $f_n$ are not very continuous themselves.
Not sure though if there is an example intermediate between the infinite limit and the nowhere continuous examples...
I'm thinking about things like $f_n(x)=\sin(\frac 1x)\times\chi_{[\frac 1n,1]}(x)$
