Let $P(x)$ be any polynomial and suppose that $a_n \rightarrow a$. Prove $lim_{n\rightarrow\infty} P(a_n) = P(a)$ I know the limit rules but that's not helping me out much here this seems so simple but I don't even know where to start. I read the chapter on this and they don't do any examples like this.  I believe a sub n converges to a.
 A: Clearly, (as you know continuity), 
 the polynomial $P$ is a continuous function mapping from $\mathbb{R}$ to $\mathbb{R}$.
Sequential continuity coincides with continuity in this case, then $a_n\to a$ implies $P(a_n)\to P(a)$.
See sequential continuity and continuity
A: OK, check (as you mentioned, by induction, or whatever method you prefer) that for all $k$, $a_n^k\to a^k$. For this, use that the limit of a product of the product of the limits.
Now, if $a_n^k\to a^k$, then also $ca_n^k\to ca^k$ for any constant $c$.
Finally, note that any polynomial is a sum of terms of that form, and use that the limit of a sum is the sum of the limits.
More carefully, if $P(x)=b+cx+dx^2+\dots+ex^m$, then $$P(a_n)=b+ca_n+da_n^2+\dots+ea_n^m,$$ and each term $ra_n^k$ converges to $ra^k$. So $P(a_n)$, being a sum of these terms, converges to $$b+ca+da^2+\dots+ea^m=P(a).$$
A: If you know the Ruffini identity, which is connected to the Horner scheme, then you know that there exists a polynomial $q(a,x)$ such that
$$
p(x)=p(a)+(x-a)q(a,x)
$$
Now you need only to argue that $q(a,x)$ is bounded if the distance from $x$ to $a$ is bounded to get the estimate
$$
|p(x)-p(a)|\le M(r)\,|x-a| \quad\text{for}\quad |x-a|<r.
$$
