Exercise 4.2.8

Question

Compute each limit or state that it does not exist. Use the tools developed in this section to justify each conclusion.
  \begin{enumerate}[label=(\alph*)] 
  \item $\lim _{x \rightarrow 2} \frac{|x-2|}{x-2}$
  \item $\lim _{x \rightarrow 7 / 4} \frac{|x-2|}{x-2}$
  \item $\lim _{x \rightarrow 0}(-1)^{[[1 / x]]}$
  \item $\lim _{x \rightarrow 0} \sqrt[3]{x}(-1)^{[[1 / x]]}$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Does not exist, the sequence $x_n = 2 + 1/n$ makes $|x-2|/(x-2)$ converge to $1$, but $x_n = 2 - 1/n$ makes $|x-2|/x-2$ converge to $-1$. ($x \to |x|$ is not differentiable at zero for the same reason)
  \item $-1$. For $\delta < 1/4$ then $x<2$ and we just have $-1$.
  \item Does not exist, let $(x_n) = 1/2n$ and $(y_n) = 1/(2n + 1)$, then clearly $\lim x_n = \lim y_n = 0$ but $\lim f(x_n) = 1 \neq \lim f(y_n) = -1$

\item $(-1)^{[[1 / x]]}$ is bounded and $\lim_{x\to 0} \sqrt[3]{x} = 0$, so by Exercise 4.2.7 $\lim _{x \rightarrow 0} \sqrt[3]{x}(-1)^{[[1 / x]]} = 0$.

We can also show this directly, since $|\sqrt[3]{x}(-1)^{[[1/x]]}| = |\sqrt[3]{x}| < \epsilon$ when $\delta=\epsilon^{3}$.
   \end{enumerate}

Exercise 4.2.8

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Exercise 4.2.8

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