= Solution
{c}
At $x=0$, $f_n(x)=0$. For fixed $x>0$,
$$
0\leq f_n(x)=\frac{nx}{1+n^2x^2}\leq\frac1{nx}\longrightarrow0.
$$
Thus the <pointwise limit> is zero everywhere on $[0,1]$. However, with $t=nx$, the inequality $1+t^2\geq2t$ gives $f_n(x)\leq1/2$, and equality occurs at $x=1/n\in[0,1]$. Hence
$$
\boxed{\|f_n\|_\infty=\frac12\quad\text{for every }n\geq1,}
$$
so there is \b[no <uniform convergence>]. The moving evaluation point $1/n$ captures a peak missed by each fixed positive input.
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