TITLE: Showing that a sequence does not converge uniformly to a piecewise function
QUESTION [0 upvotes]: I understood how to show that a sequence converge pointwise and uniformly to another function. However, I am finding it difficult to show that a function DOES NOT converge uniformly to a function, in particular when the pointwise limit is piecewise. Let me give an example:
Let $f_n(x)=\dfrac{x}{1+x^n}$ for $n\in \mathbb{N}$ and $x\in[0, \infty)$
I am required to find the pointwise limit, which I found to be:
$$
f(x)=
\begin{cases}
x \ , \ x\in[0, 1)\\
\frac{1}{2} \ , \ x=1\\
0 \ , \ x\in(1, \infty)
\end{cases}
$$
Now for the second part of the question, I am required to show that it does not converge uniformly. I understand that the reason for this (intuitively) is because there is a sort of "jump" at $x=1$
My attempt was as follows:
Let $x=1+\frac{1}{n}$
Now I try to show that for any $N\in\mathbb{N}$ , $\exists n\geq N$, such that $|f_n-f|>\epsilon$ for some $\epsilon$. 
Let $\epsilon=1/4$ and let $n>\dfrac{1}{\sqrt[N]{3}-1}$
$$|f_N(x)-f(x)|$$
$$=\left|\dfrac{x}{1+x^N}-0\right|$$
$$=\left|\dfrac{x}{1+x^N}\right|$$
$$=\left|\dfrac{1+\frac{1}{n}}{1+(1+\frac{1}{n})^N}\right|$$
$$>\dfrac{1}{1+(1+\frac{1}{n})^N}$$
$$>1/4=\epsilon$$
First of all I am not sure if this method is correct, and even if it is, is this the best way to go about showing uniform convergence, at least in this case? 
I am aware that uniform convergence occurs iff $\lim_{n\rightarrow \infty}\sup{|f_n(x)-f(x)|}=0$, but I cannot see how to use this since the pointwise limit is a piecewise function.

REPLY [1 votes]: Each $f_n$ is continuous, but $f$ isn't. Therefore, the convergence is not uniform.
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