# Continuity of composition of continuous functions

I've learned that composition of continuous functions is continuous. ##\log x## and ##|x|## are continuous functions, but it seems that ##\log |x|## is not continuous. Is this the case?

## Answers and Replies

member 587159
I've learned that composition of continuous functions is continuous. ##\log x## and ##|x|## are continuous functions, but it seems that ##\log |x|## is not continuous. Is this the case?

Why is ##\log|x|## not continuous?

Why is ##\log|x|## not continuous?
Because of the vertical asymptote at 0?

member 587159
Because of the vertical asymptote at 0?

But how can you apply the theorem you just stated if ##\log(x)## isn't even defined at ##x = 0##? If you start with false assumptions, surely you will get wrong results.

• Mr Davis 97
I've learned that composition of continuous functions is continuous
In this theorem you must also trace carefully domains of the functions. The exact formulation is as follows. Let ##X,Y,Z## be topological spaces and ##f:X\to Y,\quad g:Y\to Z## be continuous functions. Then the function ##g\circ f:X\to Z## is a continuous function.
In your case ##f(x)=|x|,\quad g(y)=\log y,\quad X=\mathbb{R}\backslash\{0\},\quad Y=\{y\in\mathbb{R}\mid y>0\},\quad Z=\mathbb{R}##

• member 587159
WWGD
Science Advisor
Gold Member
I've learned that composition of continuous functions is continuous. ##\log x## and ##|x|## are continuous functions, but it seems that ##\log |x|## is not continuous. Is this the case?

But remember that the composition may not even be defined at some points, let alone be continuous. Try specifying the theorem a bit more carefully, as Zwierz , I think, suggested. EDIT: You may want to consider the inverse image of open/closed set is open/closed definition as well.

Last edited:
mathwonk
Science Advisor
Homework Helper
2020 Award
as suggested above, it is crucial to have a complete and precise statement of a theorem, including hypotheses. in your case the theorem says roughly, if f is continuous at the point p, and if also g is continuous at (and defined on a neighborhood of) the point f(p), then gof is (defined on a neighborhood of and) continuous at the point p. try to see what the points p and f(p) are in your case that give a problem, if any.