Prove $g_n(x)=\frac{x}{1+x^n}:[0,\infty)\rightarrow\mathbb{R}$ does not converge uniformly

Question

I say the sequence of function converges pointwise to the following

1. $x$, if $0\leq x<1$
2. $\frac{1}{2}$, if $x=1$
3. 0, if otherwise

The question specifically asked me to prove why does it NOT uniformly convergent, which means $\exists\epsilon >0\forall n\in\mathbb{N}\exists x\in [0,\infty)$ such that $|g_n(x)-g(x)|\geq\epsilon$.

But I just yield contradiction in all cases, say if $x=1$, $|g_n(x)-g(x)|=\frac{1}{2}-\frac{1}{2}=0\geq\epsilon$. But $\epsilon>0$, which is impossible.

For $x>1$,

\begin{align*} \frac{x}{1+x^n}\geq\epsilon&\implies x\geq\epsilon (1+x^n) \end{align*}

But $x$ is a constant, exponential function is unbounded and this results means exponential function is bounded.

For $0\geq x<1$,

\begin{align*} |\frac{x}{1+x^n}-x|&=|1-\frac{1}{1+x^n}||x|&=|\frac{x^n}{1+x^n}||x| \end{align*}

The way I see it is that for it to be greater than a constant, the left term needs to be greater than 1, but no way can the left term be greater than 1.

Am I wrong?

Answer 1

$g_n (1+\frac1 n)-g( 1 +\frac 1 n)=\frac {1+\frac 1n} {1+(\frac 1 n)^{n}} \to \frac 1 {1+e}$ and this proves that the convergence is not uniform. [Take $\epsilon =\frac 1 {2(1+e)}$ in the definition of uniform convergence].