For $n=1,2,3,...$, and $x\in\mathbb{R}$, define $f_n(x)=\frac{x}{1+nx^2}$. show that $\{f_n\}$ converges to some function $f$ uniformly.
Proof: In order to determine what $f$ is, we let $n\rightarrow\infty$, then we have $f=0$.
(Please help me check if these steps are correct): Since $f=0$, $|f_n|=|f_n-f|$. Therefore, we can use the property of first derivative to find the maximum or minimum $f_n$, i.e.$||f_n||$. After taking the first derivative of $f_n$, we have that the maximum value occurs at $\frac{1}{\sqrt n}$. So the value of $f_n(\frac{1}{\sqrt n})$ is $\frac{1}{2\sqrt n}$. At here, we know that as long as at those extreme points we have are less than $\epsilon$ distance then every other points would certainly be less than $\epsilon$.
Thus, for all $\epsilon>0$, and choose some $N>(\frac{1}{2\epsilon})^2$. Then $\forall n\geq N$, $|f_n(x)-f(x)|<\epsilon$.
Is the proof correct?
hint
$f_n $ is an odd function.
for $x\ge 0$, put
$$g_n (x)=|f_n (x)-0|=\frac {x}{1+nx^2} $$
$$g_n'(x)=\frac {1-nx^2}{(1+nx^2)^2}$$
$$\lim_{n\to+\infty}g_n(\frac {1}{\sqrt {n}})=0$$