Prove that $f_n(x)=\frac{1-(x/b)^n}{1+(a/x)^n}$ is uniformly convergent on the interval $[a+\epsilon ; b - \epsilon]$

Question

Title says it all; I have to prove that the function sequence $f_n(x)=\dfrac{1-(x/b)^n}{1+(a/x)^n}$ is uniformly convergent on the interval $[a+\epsilon ; b - \epsilon]$, with $0<\epsilon<(b-a)/2$ - I've already shown that the sequence is pointwise convergent on $f(x)=1$, and intuitively it's pretty obvious, that for a high enough $N \in \mathbb{N}$ that $\sup\{|f_n(x)-1|\}<\epsilon$ for $n \geq N$ - but I am having trouble actually proving it, that is writing down a proper proof of it. Any hints?

Much appreciated

Answer 1

For all $x \in [a - \epsilon, b + \epsilon]$,

$$|f_n(x) - 1| = \frac{(a/x)^n + (x/b)^n}{1 + (a/x)^n} \le (a/x)^n + (x/b)^n \le \left(\frac{a}{a + \epsilon}\right)^n + \left(\frac{b - \epsilon}{b}\right)^n.$$

Thus

$$\sup_{x\in [a-\epsilon,b+\epsilon]} |f_n(x) - 1| \le \left(\frac{a}{a+\epsilon}\right)^n + \left(\frac{b-\epsilon}{b}\right)^n.$$

Now finish the argument.