How to calculate $\limsup_{n \rightarrow \infty} |\frac{x^n}{1+x^n}|$

Question

I would like to show that the sequence $f_n = \frac{x^n}{1+x^n}$ is uniformly convergent on the interval $[0,1)$.

I know that the limiting function f of the sequence on this interval is $0$.

I would like to use the fact that $f_n$ is uniformly convergent if $\limsup_{n \rightarrow \infty} |f_n(x) - f(x)| = 0$.

This leaves me needing to show that:

$\limsup_{n \rightarrow \infty} |\frac{x^n}{1+x^n}| = 0$

Intuitively i can see that the limit of the supremum of the sequence tends to $0$ as n tends to infinity. However, I am not sure how to prove this.

Is this the right approach or is there another way to show uniform convergence that is simpler and doesn't involve the limsup.

Thanks

Answer 1

For each $n\in\Bbb N$,$$\sup_{x\in[0,1)}\frac{x^n}{1+x^n}=\frac12,$$since you always have $\frac{x^n}{1+x^n}<\frac12$ and $\lim_{x\to1}\frac{x^n}{1+x^n}=\frac12$. So,$$\lim_{n\to\infty}\sup_{x\in[0,1)}\frac{x^n}{1+x^n}=\frac12.$$From this, together with the fact that$$\bigl(\forall x\in[0,1)\bigr):\lim_{n\to\infty}\frac{x^n}{1+x^n}=0,$$you can deduce that your sequence dos not converge uniformly.