Show that $f_n(x)=\frac{1-(x/b)^n}{1+(a/x)^n}$ with $0<a<b,$ and $x \in [a,b]$ is not uniformly convergent

Question

So, I have to show that $f_n(x)=\dfrac{1-(x/b)^n}{1+(a/x)^n}$ with $0<a<b$ and $x \in [a,b]$ is pointwise convergent, but not uniformly convergent. The pointwise convergence is pretty straight forward, I get $f_n(x) \to f(x)$ with:

$f(x) = \left\{ \begin{array}{llllllr} 1/2 &,& x=a \\ 1 &,& x \in (a,b) \\ 0 &,& x=b \end{array} \right.$

And intuitively, it's pretty obvious, that function sequence is not uniformly convergent, but I'm not quite sure how to show it. I'm thinking, that I have to actually show it by definition, that is:

$\exists \epsilon>0 \forall N \in \mathbb{N} \exists n \geq N \exists x \in [a,b]: n \geq N \Rightarrow \sup\{|f_n(x)-f(x)|\} \geq \epsilon$

But I am quite stuck, to be honest. Anyone got a hint that might get me going?

Much appreciated.

EDIT: Actually I just realized, that I can just go by the fact, that if a sequence of continuous functions converges uniformly, it will converge to a continuous function - since it does not, the convergence is not uniform. That should do, right?

Answer 1

A sequence of continuous functions converging uniformly will converge to a continuous function. So assume we had uniform convergence...