Proving Pointwise Convergence of $f_n(x) = \frac{x}{1 + x^n}$

Question

I am currently trying to show that the sequence of functions defined by $f_n(x) = \frac{x}{1 + x^n}$ converges pointwise on $U = [0, \infty)$. I have found the limits for the three specific cases and they are: \begin{equation*} \lim_{n \to \infty} f_n(x) = \begin{cases} x & \text{if $x \in [0, 1)$} \\ \frac{1}{2} & \text{if $x = 1$} \\ 0 & \text{if $x \in (1, \infty)$} \end{cases} \end{equation*} I have started the proof for when $x \in [0, 1)$ but I am not sure how to proceed. Here is what I have:

Let $\epsilon > 0$. We want to show that $(f_n(x))_n$ converges to $x$ on $x \in [0, 1)$, then there exists an $N \in \mathbb{N}$ such that \begin{equation*} \left|\frac{x}{1 + x^n} - x\right| < \epsilon \end{equation*} for all $n \geq N$. Then \begin{equation*} \left|\frac{x}{1 + x^n} - x\right| = \frac{x^{n + 1}}{1 + x^n} < \epsilon \end{equation*} This is the part where I am lost, but I know that I am suppose to find an $N = N(x, \epsilon)$, however, where it confuses me is where we have the $n$ as the exponent and it is a bit unclear to me how to proceed from here. Would like some assistance for this part.

Answer 1

I will asume that $\epsilon <1$ and let you see what heppens if $\epsilon \geq 1$.

$\frac{x^{n + 1}}{1 + x^n} < \epsilon$ if $\frac{x^{n }}{1 + x^n} < \epsilon$ which is true if $x^{n} <\frac {\epsilon} {1-\epsilon}$. This is true if $n >\ln (\frac {\epsilon} {1-\epsilon})/\ln x$. [Note that $\ln x <0$. When you divide an inequality by a negative number the inequality sign gets reversed].

Answer 2

You have $x\in[0,1),$ $\epsilon>0,$ and you want $x^{n+1}<\left(1+x^n\right)\epsilon,$ i.e. $$x^n\left(x-\epsilon\right)<\epsilon.$$

• If $x\le\epsilon,$ this holds for every $n.$
• If $x>\epsilon,$ this holds iff $n\ln x<\ln\frac\epsilon{x-\epsilon},$ i.e. $n\left|\ln x\right|>\ln\frac{x-\epsilon}\epsilon,$ i.e. $$n>\frac{\ln\frac{x-\epsilon}\epsilon}{\left|\ln x\right|}.$$ For a shorter proof of the pointwise convergence in each of the three cases, see this post.