Uniformly convergent

Question

Is the sequence of function $f_1, f_2,f_3,\ldots$ on $[0,1]$ uniformly convergent if $f_n(x) = \frac{x}{1+nx^2}$? If $f_n(x) = \frac{nx}{1+n^2x^2}$?

I got the following but I think I did it wrong.

For $f_n(x) = \frac{x}{1+nx^2}$, I got if $f_n \to0$ then we must find $\epsilon>0$ an $N$ such that for $n>N$ implies $|f_n-0|<\epsilon.$ So $f_n(x) = \frac{x}{1+nx^2}$; $\lim_{n\to\infty}f_n(x) =0$. Then for $\epsilon>0$ we have $|f_n(x)-f(x)| = |\frac{x}{1+nx^2}|\le |\frac{1}{1+n}|<|\frac{1}{n}|<\epsilon$ thus $N = \frac{1}{\epsilon}$. But I think this is wrong since $|1+nx^2|<|1+n|$? How can I show it is uniformly convergent?

For $f_n(x) = \frac{nx}{1+n^2x^2}$ is not uniformly convergent. So we have $f_n \to 0$ thus $|f_n(x) - f(x)| = |\frac{nx}{1+n^2x^2}-0| = |\frac{nx}{1+n^2x^2}| \le |\frac{n}{1+n^2}|<\epsilon \implies \frac{N}{1+N^2}=\epsilon$: $N -\epsilon N^2 - \epsilon = 0$ implies $N = \frac{1\pm \sqrt{1-4\epsilon^2}}{2\epsilon}.$ $\epsilon <1/2$ thus since we cant choose any $\epsilon$ it is not uniformly convergent. Again, I think I got this wrong too?

Answer 1

*Hint For Second Problem * ${nx+{1\over nx}\over 2}\ge (nx.{1\over nx})^{1\over 2}\Rightarrow {nx\over 1+n^2x^2}\le {1\over 2}$

For the First One look that by considering $f_n'(x)=0$ we get at $x= {1\over\sqrt{n}}$, $f_n(x)$ has its supremum

so $M_n=\sup |f_n(x)-f(x)|={1\over 2\sqrt {n}}\to 0$ so by Weirstrass M test, It is uniformly convergent

Answer 2

See relevant questions on $\sum f_n(x)$ with these $f_n$'s:

Show that the series is Uniformly convergent

and

Is the sum uniformly continuous on $[0,\infty)$?

As in Marso's answer, usually with fractions it is just easy to freeze $n$ and find maximum/supremum of $f_n(x)$. then $f_n \to 0$ uniformly if these sup converge to zero with $n$.