= Solution
Put $S=\|\sigma\|_\infty$, $K=\|b\|_\infty$, and write $X_t-x_0=M_t+A_t$, where $M_t=\int_0^t\sigma(X_s)\,dB_s$ and $A_t=\int_0^tb(X_s)\,ds$. The <Itô isometry> and boundedness give
$$
\mathbb E|X_t-x_0|^2\le2S^2t+2K^2t^2\longrightarrow0.
$$
Continuity and boundedness of $b$ then imply $\mathbb E b(X_s)\to b(x_0)$. Since $M$ has mean zero,
$$
\frac{\mathbb EX_t-x_0}{t}=\frac1t\int_0^t\mathbb E b(X_s)\,ds\longrightarrow b(x_0).
$$
For the <variance>, $\mathbb EM_t^2=\int_0^t\mathbb E\sigma(X_s)^2\,ds$. The remaining terms satisfy
$$
\mathbb EA_t^2\le K^2t^2,\qquad |\mathbb E(M_tA_t)|\le SKt^{3/2},\qquad (\mathbb E(X_t-x_0))^2\le K^2t^2.
$$
Continuity of $\sigma$ and the initial $L^2$ convergence give $\mathbb E\sigma(X_s)^2\to\sigma(x_0)^2$. Expanding the <variance> and dividing by $t$ proves the <initial mean and variance derivatives of an Itô diffusion>:
$$
\boxed{\left.\frac d{dt}\mathbb EX_t\right|_{0+}=b(x_0),\qquad\left.\frac d{dt}\operatorname{Var}(X_t)\right|_{0+}=\sigma(x_0)^2.}
$$
The derivatives at the initial time are right derivatives because the process is indexed by $t\ge0$.
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