Proving that the sequence of function isn't uniformly convergent. 
Let $$f_{n}(x)=\frac{1+x}{1+\exp(nx)},\qquad n\in \mathbb{N}$$ be defined on $\mathbb{R}$. Prove that this is not uniformly convergent on the interval, that includes zero.

Without proving it, first we see that
$$
f(x)=\lim_{n\to\infty}\frac{1+x}{1+\exp(nx)}=\begin{cases}
0 & \text{ if } x>0 \\ 
\frac{1}{2} & \text{ if } x=0 \\ 
1+x & \text{ if } x<0
\end{cases}
$$
so the pointwise limit $f(x)$ is continous for all $x\in \mathbb{R}\setminus \lbrace 0\rbrace$. That is, it is not continous on $0$. Thus $\lbrace f_{n}\rbrace$ is not uniformly convergent on $\mathbb{R}$. This is what I have read about the uniformly convergene of the sequence of function in a Theorem as shown below (translated from Norwegian)

Let $f$ and $f_{1},f_{2},f_{3},\dots$ be function defined on $A$. Suppose that $f_{1},f_{2},f_{3},\dots$ are continous and that $\lbrace f_{n}\rbrace$ uniformly converges to $f$ on $A$. Then $f$ is continous on $A$.

Now I have to prove it (look at the bold text) - which is my problem I don't know how and where to start. I think I would prove it by a contradiction. Can you help me? 
Note that I don't have to prove the theorem. I only need to prove that this sequence of function isn't uniformly convergent.
 A: The theorem you mention in your question is classical result,
and the proof is actually quite simple (see the uniform limit theorem).
Even if you don't end up using it,
I really recommend reading it,
as it makes use of a trick you can use in many proofs regarding convergence of functions.
Now,
using this theorem to prove what you want is easy:
As you mention in one of your comments you can use the contrapositive of the result:

If $f$ is not continuous,
  then $f_n$ does not converge uniformly or there exists $n\in\mathbb N$ such that $f_n$ is not continuous.

Since $f_n$ is continuous for all $n$,
the result immediately follows.

If you don't want to use the theorem,
I would proceed along those lines:
Recall that $g_n$ converges uniformly to $g$ is and only if $\|g_n-g\|_{\infty}\to0$,
where $\|h\|_{\infty}=\sup\{h(x):x\in I\}$ for every function $h:I\subset \mathbb R\to\mathbb R$.
If you've studied uniform convergence,
you've probably seen that if $f_n$ converges uniformly to some function $f$,
then $f_n$ converges pointwise to the same function $f$.
Therefore,
since pointwise limits are unique,
if $f_n$ converges uniformly,
it must do so towards the piecewise function $f$ you've introduced in your question.
Thus,
to prove that $f_n$ doesn't converge uniformly at all,
all we need to do is prove that it doesn't converge uniformly to $f$.
So let $(a,b)\subset\mathbb R$ be an interval that includes zero,
and let $\epsilon>0$ be arbitrary.
Fix $n\in\mathbb N$.
Since $f_n$ is continuous and $f_n(0)=1/2$,
then for every $\epsilon>0$,
there exists $\delta_n>0$ such that if $|x-0|<\delta_n$,
then $|f_n(x)-f_n(0)|=|f_n(x)-1/2|<\epsilon$,
or,
in other words,
$f_n(x)\in(1/2-\epsilon,1/2+\epsilon)$,
which implies that $f_n(x)>1/2-\epsilon$.
Let $c_n=\min\{\delta_n,b\}$.
Then,
we have that $c_n>0$ (since $\delta_n,b>0$),
and that for any $x\in(0,c_n)$,
\begin{align*}
|f_n(x)-f(x)|&=|f_n(x)-0|&\text{(since $x>0$)}\\
&=f_n(x)&\text{(since $f_n(x)>0$ if $x\geq0$)}\\
&>\epsilon-1/2.
\end{align*}
Therefore,
since $n\in\mathbb N$ was arbitrary,
we conclude that
$$\|f_n-f\|_{\infty}\geq\epsilon-1/2\text{ for every }n\in\mathbb N.$$
If you pick $\epsilon<1/2$,
one can easily see that this implies the desired result.
A: $f_{n}$ is continuous at $x=0$ while its limit "f(x)" is not. Therefore the convergence can not be uniform.
