Solutions

(a) As a reminder, polynomials are of the form

$$a_{n}x^{n} + a_{n - 1}x^{n - 1} + \dots + a_2x^2 + a_1x + a_0.$$

Therefore, if the sequence $x_{n}$ converges to $x$, we can use the algebraic theorem's rules to prove that $p(x_{n}) \rightarrow p(x)$. That is, for each individual term of a given polynomial, we apply the rules (i) and (iii) to obtain their individual limits. Next, we apply rule (ii) to the whole sum, given our previously computed individual limits.

(b) Consider the floor function

$$f(n) = \lfloor n \rfloor$$

presented in the previous subsection. Consider also the following sequence converges ever so slowly to 1:

$$a_{n} = (0, \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \dots).$$

We know that $a_{n}$ converges to 1 and also we have $f(1) = 1$. However, $f(a_{n})$ does not converge to $1$ as well. Instead,

$$f(a_{n}) \rightarrow 0.$$