uniform convergence of $\left(\frac{x^2}{1+x^2} \right)^n$

Question

I want to find whether $f_n(x)=\left(\frac{x^2}{1+x^2} \right)^n$ on $[0,\infty)$ uniformly converges or not.

First after rearranging, I have \begin{align} \left( \frac{x^2}{1+x^2} \right)^n = \frac{1}{(1+\frac{1}{x^2})^n} \end{align} First I know $1+\frac{1}{x^2} >1$ for $x \geq 0$, and in this case $f_n \rightarrow 0$ as $n\rightarrow \infty$. [Actually I have trouble with $x \rightarrow \infty$, in this case $1+\frac{1}{x^2} =1$ and so $f_n \rightarrow 1$]

Now I am trying to show whether $f_n \rightrightarrows f=0$ To do that, I am trying to show $\lim_{n\rightarrow \infty} T_n =0$ where \begin{align} T_n = \sup\{|f_n(x) - f(x)| | x\in D\} \end{align} From binomial expansion of $(1+\frac{1}{x^2})^n$, I have \begin{align} \left( \frac{x^2}{1+x^2} \right)^n = \frac{1}{(1+\frac{1}{x^2})^n} \leq \frac{1}{1+\frac{n}{x^2}} < 1 \end{align} This seems $\lim_{n \rightarrow \infty} T_n \neq 0$...

Am I doing wrong?

Answer 1

Sketch: Pointwise, we see that \begin{align} \lim_{n\rightarrow \infty}\left(\frac{x^2}{1+x^2} \right)^n = 0. \end{align} Hence, if $f_n\rightarrow f$ uniformly then $f \equiv 0$, i.e. for every $\varepsilon>0$ there exists $N$ such that for $n\ge N$, we have that \begin{align} |f_n(x)-f(x)| = |f_n(x)|<\varepsilon \end{align} for all $x \in [0, \infty)$. However, consider $x_n = n$, then we have that \begin{align} |f_n(n)|<\varepsilon \end{align} which is false (Why?).