Exercise 4.3.4

Question

Assume $f$ and $g$ are defined on all of $\mathbf{R}$ and that $\lim_{x \rightarrow p} f(x)=q$ and $\lim _{x \rightarrow q} g(x)=r$.
  \begin{enumerate}[label=(\alph*)] 
  \item Give an example to show that it may not be true that
    $$
    \lim _{x \rightarrow p} g(f(x))=r
    $$
  \item Show that the result in (a) does follow if we assume $f$ and $g$ are continuous.
  \item Does the result in (a) hold if we only assume $f$ is continuous? How about if we only assume that $g$ is continuous?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Let $f(x) = q$ be constant and define $g(x)$ as
    $$g(x) = \begin{cases}(r/q)x &	ext{ if } x 
e q \ 0 &	ext{ if } x = q \end{cases}$$
    We have $\lim_{x 	o q} g(x) = r$ but $\lim_{x 	o p} g(f(x)) = g(q) = 0$.

The problem is that functional limits allow jump discontinuities by requiring $y 
e q$ in $\lim_{y 	o q} g(y)$ but $f(x)$ might not respect $f(x) 
e q$ as $x 	o p$.
    Continuity fixes this by requiring $\lim_{y 	o q} g(y) = g(q)$ so that $f(x) = q$ doesn't break anything.

Another fix would be requiring $f(x) 
e q$ for all $x 
e p$ - In other words that the error is always greater then zero $0 < |f(x) - q| < \epsilon$ similar to $0<|x-p|<\delta$. This would allow chaining of functional limits, however it would make it impossible to take limits of ``locally flat'' functions.
  \item Theorem 4.3.9 (Proved in Exercise 4.3.3)
  \item Not if $f$ is continuous (in our example $f$ was continuous). Yes if $g$ is continuous since it would get rid of the $f(x) = q$ problem.
   \end{enumerate}

Exercise 4.3.4

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Exercise 4.3.4

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