uniform convergence of sequence of functions $f_n(x)=\frac{nx}{1+n^2x^2} , \vert{x}\vert\le1$ and $g_n(x)=\frac{x}{1+nx^2} , \vert{x}\vert\le1$

Question

Consider sequence of fucntions: $f_n(x)=\frac{nx}{1+n^2x^2} , \left|{x}\right| \le1$ and $g_n(x)=\frac{x}{1+nx^2} , \left|{x}\right| \le1$

1.Sequence $f_n$ is uniformly convergent but sequence $g_n$ is not uniformly convergent

2.Sequence $f_n$ is not uniformly convergent but sequence $g_n$ is uniformly convergent

3.Neither $f_n$ nor $g_n$ converges uniformly

4.Both $f_n$ nor $g_n$ converges uniformly

sequence $f_n(x)$ converges point wise to 0 for $\left|{x}\right| \le1$. To check for uniform convergence consider $\left|{f_n(x)-f(x)}\right|=\left|{\frac{nx}{1+n^2x^2} -0}\right| \le\left|{\frac{n}{1+n^2x^2} }\right| \le{n} $ after this how to conclude?

sequence $g_n(x)$ converges point wise to 0 for $\left|{x}\right| \le1$. To prove for uniform convergence consider $\left|{g_n(x)-f(x)}\right|=\left|{\frac{x}{1+nx^2} -0}\right|\le\left|{\frac{1}{1+nx^2} }\right| \le \epsilon \forall n\ge {N} \mathrm{where} N=\frac{1-\epsilon} {\epsilon x^2}$ .hence $g_n(x)$ is unformly convergent.

please correct me if i am wrong.

Answer 1

Note that $f_n(1/n)=1/2$ for all $n$ so the sequence cannot possibly converge uniformly to zero.