= Solution
Let
$$
m_n=\min_{0\leq x\leq n^{-1/2}}f(x),
\qquad
M_n=\max_{0\leq x\leq n^{-1/2}}f(x).
$$
Continuity at zero gives $m_n,M_n\to f(0)$. Part (a), with the positive weight $ne^{-nx}$, gives
$$
m_n(1-e^{-\sqrt n})
\leq
\int_0^{1/\sqrt n}nf(x)e^{-nx}\,dx
\leq
M_n(1-e^{-\sqrt n}).
$$
The <squeeze theorem> therefore yields
$$
\boxed{
\int_0^{1/\sqrt n}nf(x)e^{-nx}\,dx\to f(0)}.
$$
Since $f$ is bounded on $[0,1]$, say $|f|\leq K$, the omitted tail satisfies
$$
\left|
\int_{1/\sqrt n}^1nf(x)e^{-nx}\,dx
\right|
\leq K(e^{-\sqrt n}-e^{-n})\longrightarrow0.
$$
Adding the tail proves
$$
\boxed{\int_0^1nf(x)e^{-nx}\,dx\to f(0)}.
$$
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