2
$\begingroup$

My colleague and I are stuck on the answer provided by our textbook for the following question.

Consider the functions $\begin{equation*}f(x) = \begin{cases} x+1 & x \neq 0\\ 0 & x = 0\\ \end{cases} \end{equation*}$ and $\begin{equation*}g(x) = \begin{cases} x & x \neq 1\\ 0 & x = 1\\ \end{cases} \end{equation*}$.

Show that $f$ is discontinuous at $x=0$ and $g$ is discontinuous at $x=1$ but $f \circ g$ is continuous at $x=0$ and $g \circ f$ is continuous at $x=1$.

We are stuck on:

$f \circ g$ is continuous at $x=0$

The answer simply provides:

$\lim_{x \to 0}f(g(x))=\lim_{x \to 0}f(0)=0=f(g(0))$

We can see that $\lim_{x \to 0}g(x)=0$ and the definition of $f$ gives us $f(0)=0$.

The important bit (and reason for posting this question) …

In every reference I can find, the following rule is only said to be true if $f$ is continuous at $\lim_{x \to c} g(x)$.

$$\lim_{x \to c}f \circ g(x) = f(\lim_{x \to c} g(x))$$

But in our question $c=0$ and $\lim_{x \to 0} g(x)=0$ and $f$ is not continuous at $0$, as seen in the definition of $f$ and stated in the question.

$\endgroup$

1 Answer 1

-1
$\begingroup$

The limit for f(g(x)) is not 0 ... it is 1. By constructing the composite function manually, it is clear that f(g(x)) = x+1 everywhere except x=0 and x=1. One can then easily see that the limit must be 1. However, since f is not continuous at 0, the composite rule, as it is written in the form given, cannot be applied.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .