Exercise 4.2.2

Question

For each stated limit, find the largest possible $\delta$-neighborhood that is a proper response to the given $\epsilon$ challenge.
  \begin{enumerate}[label=(\alph*)] 
  \item $\lim _{x \rightarrow 3}(5 x-6)=9$, where $\epsilon=1$.
  \item $\lim _{x \rightarrow 4} \sqrt{x}=2$, where $\epsilon=1$.
  \item $\lim _{x \rightarrow \pi}[[x]]=3$, where $\epsilon=1$. (The function $[[x]]$ returns the greatest integer less than or equal to $x$.)
  \item $\lim _{x \rightarrow \pi}[[x]]=3$, where $\epsilon=.01$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $|(5x-6)-9| = |5x-15| = 5|x-3| < 5\delta$ implies $\delta = 1/5$ for $\epsilon=1$.
  \item Consider edge cases: We have $|\sqrt{9} - 2| = 1$ ($x$ is $5$ above) and $|\sqrt{1} - 2| = 1$ ($x$ is $3$ below) leading us to set $\delta = 3$. This $\delta$ must work since $\sqrt{x}$ is monotone.
  \item We must have $[[x]]] = 3$, since $|[[x]] - 3| = 1 \not < 1$. Therefor $\delta = \pi - 3$.

If the question was using $\le$ instead of $<$ we would want $x \in (2, 5)$ as that is the largest neighborhood with $|[[x]]-3| \le 1$. Setting $\delta = \min\{|\pi-2|, |\pi-5|\} = \pi-2$ achieves this maximum neighborhood.
  \item Since $[[x]]$ is an integer $\epsilon = .01$ is the same as saying $[[x]] = 3$. This happens precisely when $x \in (3,4)$ hence we need $\delta = \min\{|\pi-3|, |\pi-4|\} = \pi-3$.
   \end{enumerate}

Exercise 4.2.2

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Exercise 4.2.2

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