Exercise 2.3.2

Question

Using only Definition 2.2.3, prove that if $\left(x_{n}\right) \rightarrow 2$, then
  \begin{enumerate}[label=(\alph*)] 
  \item $\left(\frac{2 x_{n}-1}{3}\right) \rightarrow 1$;
  \item $\left(1 / x_{n}\right) \rightarrow 1 / 2$.
   \end{enumerate}
  (For this exercise the Algebraic Limit Theorem is off-limits, so to speak.)

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We have $\left|\frac{2}{3} x_n - \frac 43\right| = \frac 23\left|x_n - 2\right| < \epsilon$ which can always be done since $|x_n - 2|$ can be made arbitrarily small.
  \item Let $N$ be such that $|x_n - 2| < \min\{1, \epsilon\}$. Since $x_n$ is at least $1$ we can bound $|1/x_n| \le 1$, giving
    $$
    |1/x_n - 1/2| = \frac{|2-x_n|}{|2x_n|} \le \frac{|x_n-2|}{2} \le \frac{\epsilon}{2} < \epsilon.
    $$
   \end{enumerate}

Exercise 2.3.2

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Exercise 2.3.2

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