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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!

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you can directly follow your remark, note that to show $f_n$ converges pointwise on $[0,2]$, you only deal with a fixed x, so you can just seperate the cases as you mentioned above. –  ougao Dec 14 '12 at 17:59
    
you'd better use $x\in [0,2]$. –  ougao Dec 14 '12 at 17:59
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Take it on a case by case basis. If you noted that for fixed $x$ with $0\le x<1$, you have $0\le f_n(x)\le x^n$, could you then show $f_n(x)\rightarrow 0$? For $1\le x\le 2$ you might start by noting $f_n(x)={1\over \textstyle{1\over x^n}+1}$. –  David Mitra Dec 14 '12 at 18:02
    
You have already proved your result. Is the issue that you need to give an epsilon delta proof (i.e. a more detailed proof) and don't know how to do that for pointwise convergence? –  John Dec 14 '12 at 18:30

1 Answer 1

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 $x \in (1,2]$.

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