Exercise 4.2.6

Question

Decide if the following claims are true or false, and give short justifications for each conclusion.
  \begin{enumerate}[label=(\alph*)] 
  \item If a particular $\delta$ has been constructed as a suitable response to a particular $\epsilon$ challenge, then any smaller positive $\delta$ will also suffice.
  \item If $\lim _{x \rightarrow a} f(x)=L$ and $a$ happens to be in the domain of $f$, then $L=f(a)$
  \item If $\lim _{x \rightarrow a} f(x)=L$, then $\lim _{x \rightarrow a} 3[f(x)-2]^{2}=3(L-2)^{2}$
  \item If $\lim _{x \rightarrow a} f(x)=0$, then $\lim _{x \rightarrow a} f(x) g(x)=0$ for any function $g$ (with domain equal to the domain of $f$.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Obviously, since if $\delta' < \delta$ then $|x - a| < \delta'$ implies $|x - a| < \delta$.
  \item False, consider $f(0) = 1$ and $f(x) = 0$ otherwise, the definition of a functional limit requires $|x - a| < \delta$ to imply $|f(x) - L| < \epsilon$ \emph{for all $x$ not equal to $a$} (This is the $0<|x-a|$ part)
  \item True by the algebraic limit theorem for functional limits. (or composition of continuous functions, but that's unnecessary here)
  \item False, consider how $f(x) = x$ has $\lim_{x 	o 0} f(x) = 0$ but $g(x) = 1/x$ has $\lim_{n 	o 0} f(x)g(x) = 1$. (Fundementally this is because $1/x$ is not continuous at $0$)
   \end{enumerate}

Exercise 4.2.6

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Exercise 4.2.6

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