If $f_n \to f$ uniformly and $f$ has a discontinuity at $x_0$, then there exists $N$ such that $f_N$ has a discontinuity at $x_0$. I want to prove or disprove:
If $f_n \to f$ uniformly and $f$ has a discontinuity at $x_0$, then there exists $N$ such that $f_N$ has a discontinuity (of the same type) at $x_0$. 
However, I am not sure how to approach this since the discontinuity could be of any type. Is there a way to prove this without using several cases?
 A: Hint  The contrapositive of this is: If $f_n \to f$ uniformly, and each $f_n$ is continuous at $x_0$ then $f$ is continuous at $x_0$.
In this form, you don't need to worry about what type of discontinuity you have to deal with.
Or better Assume by contradiction that each $f_n$ is continuous at $0$. Prove that $f$ is continuous at $0$.
Added For the same type:
$$f_n= \left\{ 
\begin{array}{lc}
\frac{1}{n} &\mbox{ if } x >0 \\
0 & \mbox{ if }  x<0 \\
1 & \mbox{ if } x=0
\end{array}
\right.$$
converges uniformly to
$$f_n= \left\{ 
\begin{array}{lc}
0 & \mbox{ if }  x \neq 0 \\
1 & \mbox{ if } x=0
\end{array}
\right.$$
Note that $f$ has removable discontinuity while all $f_n$ have jump.
Added 2: Try proving instead the following: If $f_n \to f$ uniformly, and 
$$\lim_{x \to x_0^+} f_n(x)=l_n$$
exists for all $n$, then $\lim_{x \to x_0^+} f(x)$ exists.
To do this, show that $l_n$ is Cauchy, deduce that converges to some $l$ and show that $\lim_{x \to x_0^+} f(x)=l$.
Then, do the same for the left limit.
A: $f$ is continuous at $x_{0}$ if and only if for any sequence $(x_{n})$ such that $x_{n}\rightarrow x_{0}$, it follows that $f(x_{n})\rightarrow f(x_{0})$
Say now $f$ is discontinuous at $x_{0}$, by passing to a subsequence, there is some $\epsilon>0$ such that $|f(x_{n})-f(x_{0})|\geq\epsilon$ for $n\geq N$ for a sequence $(x_{n})$ such that $x_{n}\rightarrow x_{0}$. Then choose an $N$ such that $|f_{N}(x)-f(x)|<(1/3)\epsilon$ for all $x$. We see that
\begin{align*}
|f_{N}(x_{n})-f_{N}(x_{0})|\geq|f(x_{n})-f(x_{0})|-|f_{N}(x_{n})-f(x_{n})|-|f_{N}(x_{0})-f(x_{0})|>(1/3)\epsilon
\end{align*}
for all such $n\geq N$, so $f_{N}(x_{n})$ does not converge to $f_{N}(x_{0})$, $f_{N}$ is discontinuous at $x_{0}$.
