Exercise 2.3.1

Question

Let $x_{n} \geq 0$ for all $n \in \mathbf{N}$.
  \begin{enumerate}[label=(\alph*)] 
  \item If $\left(x_{n}\right) \rightarrow 0$, show that $\left(\sqrt{x_{n}}\right) \rightarrow 0$.
  \item If $\left(x_{n}\right) \rightarrow x$, show that $\left(\sqrt{x_{n}}\right) \rightarrow \sqrt{x}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Setting $x_n < \epsilon^2$ implies $\sqrt{x_n} < \epsilon$ (for all $n \ge N$ of course)

\item We want $|\sqrt{x_n} - \sqrt x| < \epsilon$ multiplying by $(\sqrt{x_n} + \sqrt{x})$ gives $|x_n - x| < (\sqrt{x_n} + \sqrt{x})\epsilon$ since $x_n$ is convergent, it is bounded $|x_n| \le M$ implying $\sqrt{|x_n|} \le \sqrt{M}$, multiplying gives
    $$
    |x_n - x| < \left(\sqrt{x_n} + \sqrt{x}\right)\epsilon \le \left(\sqrt{M} + \sqrt x\right)\epsilon
    $$
    Since $|x_n - x|$ can be made arbitrarily small we can make this true for some $n \ge N$. Now dividing by $\sqrt{M} + \sqrt x$ gives us
    $$|\sqrt{x_n} - \sqrt{x}| \le \frac{|x_n - x|}{\sqrt{M} + \sqrt{x}} < \epsilon$$
    therefore $|\sqrt{x_n} - \sqrt x| < \epsilon$ completing the proof.
   \end{enumerate}

Exercise 2.3.1

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Exercise 2.3.1

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