Show that for a sequence $z_k\longrightarrow z_o$ we have for all polynomials: $p(z_n)\longrightarrow p(z_0)$ I was trying to study some proprieties about series and their respective convergence and I encountered a problem that I don't really know how to tackle:
Let $z_k \longrightarrow z_0$ be a convergent sequence. Then for all polynomials
$p(z) = a_nz^n + ... +a_1z+a_0$ we have: $$p(z_n) \longrightarrow p(z_0)$$
Any help would be greatly appreciated!
 A: It just requires to know that polynomials are continous functions. If you don't know the notion of continuity, you can  easily prove that the claim holds for $z^n$ using the fact that the product of two convergent sequences is a sequence that converges to the product of the limits. Then use induction to obtain the claim. After that use sum of convergent sequences...
A: *

*$z\mapsto z$ is continuous

*the product of two continuous functions is continuous, hence continuity of $z\mapsto z^k$

*finite linear combinations of continuous functions are continuous

A: Per the so-called Horner-Ruffini scheme, you can decompose your polynomial as
$$
p(z)=p(z_0)+(z-z_0)·q(z)
$$
for some polynomial $q$ whose coefficients are also computed by the scheme.
Now you need to know that polynomials are bounded over bounded intervals to find
$$
|p(z)-p(z_0)|\le |z-z_0|·\sup_{w\in[z_0-1,z_0+1]}|q(w)|
$$
for all $z$ with $|z-z_0|<1$. This local Lipschitz continuity allows for an easy translation from limits in the arguments to limits in the values of the function.
