I want to show $$f_n(x):=\frac{x}{1+n^2x^2}$$ is uniformly convergent in $\mathbb R$ using $\epsilon-n_0$ argument.
So $|f_n-f|\leq\frac{|x|}{1+n^2|x|^2}$
I know $|f_n-f|\leq\frac1n$ since $\frac1n$ is the maximum of $f_n$. I can show this using the derivatives of $f_n$.
My question is: Can you show this or any other equation (independent of $x$) without using derivatives in any calculation?
I've tried $|f_n-f|\leq \frac1n \iff -\frac1n\leq\frac{x}{1+n^2x^2}\leq\frac1n \iff\ldots$ but I get no inequality which is obviously right.
Best Answer
Since
$$\frac x{1+n^2x^2}\xrightarrow[n\to\infty]{}0\;\;,\;\;\forall\,x\in\Bbb R$$
You want to prove that
$$\forall\,\epsilon>0\;\;\wedge\;\;\forall\,x\in\Bbb R\;\;\exists\,N_\epsilon\in\Bbb N\;\;s.t.\;\;n>N_\epsilon\implies\;\left|\frac x{1+n^2x^2}\right|<\epsilon$$
But
$$\left|\frac x{1+n^2x^2}\right|=\frac{|x|}{1+n^2x^2}<\epsilon\iff n^2>\frac1{|x|\epsilon}-\frac1{x^2}$$
Thus we can choose
$$\;N_\epsilon:=\frac1{2\epsilon}\;,\;\;\text{so}\;\;n>N_\epsilon\implies n^2>\frac1{4\epsilon^2}\ge\frac1{|x|\epsilon}-\frac1{x^2}\;$$
$$\text{Why? Check the maximum of the function}\;\;f(x)=\frac1{x\epsilon}-\frac1{x^2}$$
once for $\;x>0\;$ and once for $\;x<0\;$ .