= Solution
{c}
\b[No, not from the stated assumptions alone.] The centered slope is a weighted sum,
$$
\widehat\beta_1-\beta_1=\frac{\sum_i x_i\epsilon_i}{S_{xx}},
$$
and the weighted errors need not be identically distributed. The ordinary <central limit theorem> therefore does not apply directly.
For a concrete counterexample, for every $n\ge3$ choose
$$
x=(1,-1,0,\ldots,0),\qquad z=(1,1,-2,0,\ldots,0).
$$
These satisfy all three centring and <orthogonality> assumptions, and both explanatory columns are nonzero. Let the errors be <independent> with <Rademacher distribution>, taking values $\pm1$ with equal <probability>. Then
$$
\widehat\beta_1-\beta_1=\frac{\epsilon_1-\epsilon_2}{2},
$$
which takes values $-1,0,1$ with <probabilities> $1/4,1/2,1/4$ for every $n$. Increasing the sample size adds no information about this slope, and the standardized estimator does not approach a <normal distribution>.
A useful additional assumption for <asymptotic normality of a weighted regression slope> is that no single design value dominates:
$$
\frac{\max_i x_i^2}{\sum_i x_i^2}\longrightarrow0.
$$
Under this condition and the given finite error <variance>, the <Lindeberg-Feller central limit theorem> yields
$$
\frac{\sqrt{S_{xx}}(\widehat\beta_1-\beta_1)}{\sigma}\xrightarrow d N(0,1),
$$
justifying the approximation $N(\beta_1,\sigma^2/S_{xx})$ for the slope as well.
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