I can't understand why this sequence of functions does not have more than one pointwise limit? I am given the sequence of functions $(f_k)_{k\in\mathbb{N}}$, where
$f_k(x)= \frac{x^k}{1+x^k}$, $k\in\mathbb{N},x\in[0,\infty)$
I need to compute the pointwise limit of it. I note that it is bounded, as $$f_k(x)\leq \frac{x^k}{x^k}=1$$
Surely for $0\leq x<1$ we have $f_k(x) \rightarrow 0$ as $k \rightarrow \infty$ and for $k>1$ we have $f_k(x) \rightarrow 1$ as $k \rightarrow \infty$? I don't understand what I am missing.
 A: This sequence of functions $f_k(x)$ converges pointwise to a function $f(x)$ on $[0,\infty)$, given by:
$$f(x)=\lim_{k\to\infty}f_k(x)=\begin{cases}0 & 0\le x \lt 1 \\ \frac{1}{2} & x=1 \\ 1 & x\gt 1\end{cases}$$
Because $f_k$ are all continuous on $[0,\infty)$ and $f$ is discontinuous, it follows that those functions converge to $f$ (when $k\to\infty$) only pointwise, but not uniformly, because otherwise their limit $f$ would be continuous as well, as per Uniform limit theorem. Actually, this sequence of functions doesn't converge uniformly on any interval containing $x=1$, for the same reason.
As for the additional question you added in the comments:
Let $b\gt 1$. All the functions $f_k(x)=\frac{x^k}{1+x^k}\lt 1$ are increasing in $x$. Thus, if, for given $\varepsilon\gt 0$ we can find $k_0\in\mathbb N$ such that $|1-f_k(b)|=1-f_k(b)\lt\varepsilon$ for all $k\gt k_0$ (which we can, because of the pointwise convergence at $b$), then, for all $x\ge b$, we will have $|1-f_k(x)|=1-f_k(x)\le 1-f_k(b)\lt \varepsilon$ for all $k\gt k_0$ too, which means that the choice of $k_0$ is independent on $x$, and so the convergence of $f_k$ to $f$ when $k\to\infty$ is uniform on any interval $[b,\infty)$ for $b\gt 1$.
A: These functions do not converge uniformly, so their pointwise limit can be discontinuous.
