= Solution
{c}
The functions $g_n$ are nonnegative. The <change of variables> $u=nx$ and the given <Gaussian integral> imply
$$
M_n:=\int_0^1g_n(x)\,dx=\frac2{\sqrt\pi}\int_0^n e^{-u^2}\,du\longrightarrow1,
\qquad 0\leq M_n\leq1.
$$
For every fixed $\delta>0$,
$$
\int_\delta^1g_n(x)\,dx\leq\frac2{\sqrt\pi}\int_{n\delta}^\infty e^{-u^2}\,du\longrightarrow0.
$$
Thus the mass concentrates at the endpoint zero. This is a <one-sided approximate identity>.
Because $f$ is <continuous> on the <compact interval> $[0,1]$, it is a <bounded function>; put $B=\sup_{[0,1]}|f|$. Given $\varepsilon>0$, choose $\delta\in(0,1)$ with $|f(x)-f(0)|<\varepsilon$ for $0\leq x<\delta$. Splitting the integral near and away from zero yields
$$
\begin{aligned}
\left|\int_0^1 f(x)g_n(x)\,dx-f(0)\right|
&\leq\int_0^1|f(x)-f(0)|g_n(x)\,dx+|f(0)|\,|M_n-1|\\
&\leq\varepsilon M_n+2B\int_\delta^1g_n(x)\,dx+|f(0)|\,|M_n-1|.
\end{aligned}
$$
Taking the <limit superior> gives a bound of $\varepsilon$, and $\varepsilon$ was arbitrary. Consequently
$$
\boxed{\int_0^1 f(x)g_n(x)\,dx\longrightarrow f(0).}
$$
Only <continuity> at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed <continuity> everywhere also guarantees the ordinary integrals exist.
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