Pointwise Convergence Problem Let  $f_{n}\left( x\right) =\dfrac {x^{n}} {1+x^{n}}$, $x \in \left[ 0,2\right]$
Show that $f_{n}$ converges pointwise on $[0,2]$.  
I know that the function converges to $0$ for $0\leq x < 1$, converges to $1/2$ for $x=1$ and converges to $1$ for $x>1$, but I need help showing the definition of pointwise convergence given $\varepsilon>0$.
Thanks!
 A: As you have correctly identified, you have that $$\lim_{n \to \infty} \dfrac{x^n}{1+x^n} = \begin{cases} 0 & x \in [0,1)\\ \dfrac12 & x = 1\\ 1 & x \in (1,2]\end{cases}$$
The proof follows immediately once you identify what the function converges to.
First let us consider the domain $[0,1)$. For $x=0$, it is trivial that the limit is indeed $0$. For $x \in (0,1)$, given $\epsilon > 0$, we can choose $N(\epsilon;x) = \left\lceil \dfrac{\log(\epsilon)}{\log(x)}\right\rceil$. Then for $n > N(\epsilon)$, we have that
$$\left \vert \dfrac{x^n}{1+x^n} - 0\right \vert \leq \left \vert x^n \right \vert < \epsilon$$ Hence, $f_n(x) \to 0$ for $x \in [0,1)$.
Now let us consider $x=1$. It is trivial that the limit is $\dfrac12$, since $f_n(1) = \dfrac12$ for all $n \in \mathbb{N}$.
Now let us consider the domain $(1,2]$. The claim is that the limit is $1$. For $x \in (1,2]$, given $\epsilon > 0$, we can choose $N(\epsilon;x) = \left\lceil -\dfrac{\log(\epsilon)}{\log(x)}\right\rceil$. Then for $n > N(\epsilon)$, we have that
$$\left \vert \dfrac{x^n}{1+x^n} - 1\right \vert = \left \vert \dfrac1{1+x^n}\right \vert  \leq \left \vert \dfrac1{x^n} \right \vert < \epsilon$$ Hence, $f_n(x) \to 1$ for $x \in (1,2]$.
