Sequence of continuous functions whose pointwise limit has a second-kind discontinuity My try was 
$$f_n = \sqrt{x-\frac{1}{2^n}}$$
such that the discontinuity of the second kind occurs at $x=0$. 
However, I imagine there's a better solution that actually matches the spirit of the question.
 A: The function $f:\mathbb{R} \rightarrow \mathbb{R}$ where
$$f(x) = \begin{cases} \sin\left(1/x\right) &\mbox{if}\,\,x \neq 0, \\ 0 &\mbox{if} \,\,x=0.\end{cases}$$
has a type II discontinuity at $x=0$, because $\lim_{x \rightarrow 0}\sin(1/x)$ does not exist.
If $x_n = (\pi/2 + 2n\pi)^{-1}$ then $x_n \rightarrow 0$ and $f(x_n) \rightarrow 1$ as $n \rightarrow \infty$.
If $y_n = (2n\pi)^{-1}$ then $y_n \rightarrow 0$ and $f(y_n) \rightarrow 0$ as $n \rightarrow \infty$.
Now consider the sequence of functions $(f_n)$ where
$$f_n(x) = \begin{cases} \sin\left[x^{-1}e^{-1/(nx^2)}\right] &\mbox{if}\,\,x \neq 0, \\ 0 &\mbox{if} \,\,x=0.\end{cases}$$
Then $f_n$ is continuous on $\mathbb{R}$ since
$$\lim_{x \rightarrow 0}\sin\left[x^{-1}e^{-1/(nx^2)}\right]=\sin(0) = 0,$$
and the sequence $(f_n(x))$ converges pointwise to $f(x)$ :
$$\lim_{n \rightarrow \infty}f_n(x)=\begin{cases} \lim_{n \rightarrow \infty}\sin\left[x^{-1}e^{-1/(nx^2)}\right]= \sin(1/x)&\mbox{if} \,\, x \neq 0,\\ \lim_{n \rightarrow \infty}(0) = 0 &\mbox{if} \,\, x = 0,\end{cases}=f(x).$$
A: How about
$$
f_n(x)=\frac{x}{x^2+1/n}?
$$
