Exercise 2.3.8

Question

Let $\left(x_{n}\right) \rightarrow x$ and let $p(x)$ be a polynomial.
  \begin{enumerate}[label=(\alph*)] 
  \item Show $p\left(x_{n}\right) \rightarrow p(x)$.
  \item Find an example of a function $f(x)$ and a convergent sequence $\left(x_{n}\right) \rightarrow x$ where the sequence $f\left(x_{n}\right)$ converges, but not to $f(x)$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Applying the algebraic limit theorem multiple times gives $(x_n^d) 	o x^d$ meaning
    $$
    \lim p(x_n) = \lim \left(a_{d}x_n^d + a_{d-1}x_n^{d-1} + \dots + a_0ight) = a_d x^d + a_{d-1}x^{d-1} + \dots + a_0 = p(x).
    $$
    As a cute corollary, any continuous function $f$ has $\lim f(x_n) = f(x)$ since polynomials can approximate continuous functions arbitrarily well by the Weierstrass approximation theorem.
  \item Let $(x_n) = 1/n$ and define $f$ as
    $$
    f(x) = \begin{cases}
      0 &	ext{if } x = 0 \
      1 &	ext{otherwise}
    \end{cases}
    $$
    We have $f(1/n) = 1$ for all $n$, meaning $\lim f(1/n) = 1$ but $f(0) = 0$.
   \end{enumerate}

Exercise 2.3.8

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Exercise 2.3.8

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