Let there be a series such that $\{ x_n \} \rightarrow x$ and $p(x)$ be a polynomial.
Show $p(x_n)$ $\rightarrow$ $p(x)$.
Attempt at solution:
Let $p(x)= a_nx^m+a_{n-1}x^{m-1} + \cdots + a_1 x+ a_o$.
Since $(x_n) \rightarrow x$ we have $n \ge N_0 \Rightarrow |x_n-x|< \epsilon$.
Also since $\lim(a_n)=a$ $\ $implies $c \lim(a_n)= ca$ it follows for (1) $n\ge N_0$ implies $|ca_n-ca|\lt \epsilon$.
Using (1) we choose $n\ge N_1 \Rightarrow |a_nx^m -a_n(x_n)^m|< \frac{\epsilon}{n} $, and $n\ge N_2 \Rightarrow |a_{n-1}x^{m-1} -a_{n-1}(x_n)^{m-1})|< \frac{\epsilon}{n}$, ..., and choose $n\ge N_n$ $\Rightarrow |a_1x^{1} -a_1(x_n)^{1})|< \frac{\epsilon}{n}$
$\ $
Now taking $n \ge M \ge \max\{N_1,N_2,..., N_n\}$ $\Rightarrow$ $|p(x_n)-p(x)|$ equals to $$ |(a_nx^m+a_{n-1}x^{m-1} + \cdots + a_1 x+ a_o) - (a_n(x_n)^m+a_{n-1}(x_n)^{m-1} + \cdots + a_1 (x_n)+ a_o)|$$
$$\le |a_nx^m -a_n(x_n)^m|+ |a_{n-1}x^{m-1} -a_{n-1}(x_n)^{m-1}|+ \cdots + |a_1 x- a_1 (x_n)^1|+|a_0-a_0|$$
$$\lt (\frac{\epsilon}{n} + \frac{\epsilon}{n}+ \cdots + \frac{\epsilon}{n}) = n\frac{\epsilon}{n} = \epsilon$$
Question: Is my solution correct? If not where have I made mistakes and can someone provide the correct proof? And is there anything I can do write a "neater" or better proof? Thanks in advance guys.
Your solution seems right, but here's a much more elegant and short proof.
Recall by definition $f$ is continuous if and only if for every $x_n \rightarrow x$, $f(x_n) \rightarrow f(x)$.
It is clear by the equivalent $\epsilon - \delta$ definition that any constant $c$ is also continuous.
In addition, we claim $x$ is continuous, which follows from letting $\delta = \epsilon$.
Then let $f(x) =\sum_{k=1}^{N} a_k x^k$ be a polynomial, then we see that we can form $f(x)$ by multiplying and adding constants and $x$. Hence, as finite products and additions of continuous functions are continuous, we see $f$ is continuous. Hence, by definition for all $x_n \rightarrow x$, $f(x_n) \rightarrow f(x)$.