Proving that $\frac{x^n}{1+x^n}$ converges simply but not uniformly We have that
$$\lim_{n\to \infty} \frac{x^n}{1+x^n} = 1$$
I want to show that for every $\epsilon>0$, there exists $n_0$ such that
$$n>n_0\implies \left|\frac{x^n}{1+x^n}-1\right| = \left|\frac{1}{1+x^n}\right|<\epsilon$$
I did:
$$n_0 =\log_{x}\frac{1}{\epsilon} $$
so
$$n>n_0\implies x^n>x^{\log_{x}\frac{1}{\epsilon}} = \frac{1}{\epsilon}$$
therefore:
$$\frac{1}{1+x^n}<\frac{1}{x^n}<\epsilon$$
Am I right?
Now, I need to show that this convergence is not uniform. I need to show that there exists an $\epsilon>0$ such that for 
$$n>n_0 \implies \left|\frac{1}{1+x^n}-1\right|>\epsilon$$
but I'm unable to do it
 A: The pointwise limit $f$ is defined by
$f(x)=1 $ if $|x|>1,$ 
$ f(1)= \frac{1}{2}$ and
$f(x)=0$ if $|x|<1$.
but
$\forall n\geq 0 \;x\mapsto \frac{x^n}{1+x^n}$ is continuous at $x=1$
and $f$ is not continuous at $x=1$
thus, the convergence is not uniform at $(-\infty,-1)\cup(-1,+\infty)$.
A: Clearly there is a problem with domain. Hopefully the following will help.
Pointwise convergence on $D \subset \mathbb R$: 
$$\forall x \in D \forall \varepsilon >0 \exists n_0(\varepsilon, x) \in \mathbb{N}\forall n \geq n_0(\varepsilon, x) \implies |f_n(x) - f(x)| \leq \varepsilon,$$
Negation of Pointwise convergence on $D \subset \mathbb R$:
$$\exists x \in D \exists \varepsilon >0 \forall n_0 \in \mathbb{N}\exists n(n_0) \geq n_0 \implies |f_n(x) - f(x)| \leq \varepsilon,$$
Uniform convergence on $D \subset \mathbb R$: 
$$\forall \varepsilon >0 \exists n_0(\varepsilon) \in \mathbb{N}\forall n \geq n_0(\varepsilon)\forall x \in D \implies |f_n(x) - f(x)| \leq \varepsilon,$$
Negation of Uniform convergence on $D \subset \mathbb R$:
$$\exists \varepsilon >0 \forall n_0 \in \mathbb{N}\exists n(n_0) \geq n_0\exists x(n_0) \in D \implies |f_n(x) - f(x)| \leq \varepsilon.$$
When I write $n_0(\varepsilon, x)$ I mean that $n_0$ depends on $\varepsilon, x$. Hope it helps.
A: I guess that you need to restrict your domain of convergence of $x$ to be $(1,\infty)$ to deduce that pointwise convergence to be $1$ for each such $x$. If you restrict such $x$ to be $[0,1)$, then the pointwise convergence is $0$. For such $x\in(1,\infty)$, the convergence is not uniform. Assuming the contrary that it is, then choose some $N\in{\bf{N}}$ such that for all $n\geq N$ and $x\in(1,\infty)$, then $\left|\dfrac{x^{n}}{1+x^{n}}-1\right|<\dfrac{1}{3}$. We have then $\dfrac{1}{1+x^{N}}<\dfrac{1}{3}$ for all $x\in(1,\infty)$. Taking $x\downarrow 1$, then $\dfrac{1}{2}<\dfrac{1}{3}$, which is of course absurd.
