Does the monotone pointwise limit of continuous functions has left and right limit everywhere? Let $\{f_n(x)\}_{n\geq 0}$ be a sequence of continuous functions such that $f_n(x) > f_{n + 1}(x)$ for all $n\geq 0$. Assume that $f_n(x)$ converges pointwise to a function $f(x)$. It is clear that $f$ is not necesarilly continuous, but, does it have right and left limit everywhere?
 A: Here is an example of such a sequence defined on $[0,1]$, where the limit as $x \to 0$ does not exist. To simplify notation, and not calculate all the lines I will use explicitly, I will denote by $d_{(x_1,y_1,x_2,y_2)}(x)$ the equation for the line passing through $(x_1,y_1)$ and $(x_2,y_2)$.
Choose your favorite converging series with only positive terms. I will write it down as $S=\sum_{k=1}^\infty a_n$. Let $S_n=\sum_{k=1}^n a_n$ be the sequence of partial sums associated to this series.
For a given integer $n$, $f_n(x)$ will be $S_n$ if $x \in [0,\frac{1}{n+1}]$. If $x \in (\frac{1}{k+1},\frac{\frac{1}{k}+\frac{1}{k+1}}{2}]$ for some $k=1,2,...,n$, we let
$$f_n(x)=d_{(\frac{1}{k+1},0,\frac{\frac{1}{k}+\frac{1}{k+1}}{2},1)}(x)+S_n.$$
If $x \in (\frac{\frac{1}{k}+\frac{1}{k+1}}{2},\frac{1}{k}]$ for some $k=1,2,..,n$, we let
$$f_n(x)=d_{(\frac{\frac{1}{k}+\frac{1}{k+1}}{2},1,\frac{1}{k},0)}(x)+S_n.$$
This defines our function. I will let you verify that it respects all your criteria.
Denote by $f$ the pointwise limit of $f_n$. Notice that for all positive integers $j$, we have that $f(\frac{1}{j})=S$   and $f(\frac{\frac{1}{j}+\frac{1}{j+1}}{2})=1+S$. This gives us two sequences that tend to $0$ with $f$ having a different value for each of the sequences, and so the limit as $x$ goes to $0$ of $f$ does not exist!
If anything needs clarification, let me know! This was a fun problem to work on, thank you for posting the question.
