= Solution
{c}
\b[Yes.] By <uniform convergence>, choose $N$ so that $|p_n(x)-f(x)|<1$ for every $n\geq N$ and every $x\in\mathbb R$. Then
$$
|p_n(x)-p_N(x)|<2\qquad(n\geq N,\ x\in\mathbb R).
$$
The <polynomial> $p_n-p_N$ is bounded on the whole <real line>. The supplied growth property of nonconstant <polynomials> implies that this difference must be constant; write $p_n=p_N+c_n$. Evaluating at zero gives $c_n=p_n(0)-p_N(0)\to c=f(0)-p_N(0)$. For each $x$, take the <limit of a sequence> in $p_n(x)=p_N(x)+c_n$ to obtain
$$
\boxed{f(x)=p_N(x)+c\quad\text{for every }x\in\mathbb R.}
$$
This is a <polynomial>. The <uniform limits of polynomials on the real line> therefore have a tail in which all nonconstant coefficients have already stabilized; only the constant coefficients may continue to change.
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