Proving uniform convergence of $f_n(x)=\frac{x^n}{(1+x^n)}$ on $[0,b]$ for $b\in (0,1)$

Question

I must prove uniform convergence of $f_n(x)=\frac{x^n}{(1+x^n)}$ on on $[0,b]$ for $b\in (0,1)$ - I have already proven that it is not uniformly convergent at 0 and 1. How should I do this? I understand that I must find $\epsilon > 0$ such that $\vert f_n(x) - L \vert < \epsilon$ for $n>n_0(\epsilon)$.

Answer 1

Let $f(x)=\lim_{n\to \infty} f_n(x)$.

Then $f(0)=0;f(x)=0;0<x<1$ So $f(x)=0$ .

Now $\sup_{x\in [0,b]}|f_n(x)|=\sup_{x\in [0,b]}\dfrac{x^n}{1+x^n}=\dfrac{b^n}{1+b^n}$ (Since $\dfrac{x}{1+x}$ is an increasing function.)

And $\dfrac{b^n}{1+b^n}\to 0(n\to \infty) $ as $b<1$ .

$f_n\to f$(uniformly ) .

Note:If $b=1$ then $\sup_{x\in [0,b]}|f_n(x)|=\dfrac{1}{2}\neq0 $

Answer 2

Another way to see this without having to calculate the limit (which in this case is easy but generally might not be) is the following.

Since the sequence $(a_n)$, $a_n = b^n$ converges to zero $\forall \epsilon >0 , \; \exists N >0 , \; \forall n>N, \; b^n< \epsilon$. We know that a sequence converges uniformly if-f it is uniformly a cauchy sequence. Therefore we have: $$\forall \epsilon >0 \; \exists n_0 = N, \; \forall n \geq m \geq N , \; \forall x \in (0,b)$$ $$ |f_n(x) - f_m(x)| = \left|\frac{x^n}{1 + x^n} - \frac{x^m}{1 + x^m}\right| = \left| \frac{x^n - x^m }{(1 + x^n)(1 + x^m)} \right| \leq \left| \frac{b^n - 0 }{(1 + 0)(1 + 0)} \right| = b^n < \epsilon $$ (the second to last inequality comes from $0<x<b$)