# Thread: Composite functions continuity theorem

1. ## Composite functions continuity theorem

g: R->R

g(x) = 2 if x not equals 1
g(x) = 0 if x equals 1

f(x) = x +1 for R

"Verifiy" (sorry my english) that lim x-> 0 (gof)(x) is not equal lim x->0 (gof)(0)

If there is a contradiction with the continuity theorem of composite function, give a justification why.

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Firs part:

lim x->0 (gof)(x) = lim x->0 g(x+1) = g(1) = 2
lim x->0(gof)(0) = g(0+1) lim x->0 g(1) = 0

??

and now what is continuity theorem of composite function?

2. ## Re: Composite functions continuity theorem

Originally Posted by Fabio010
g: R->R
g(x) = 2 if x not equals 1
g(x) = 0 if x equals 1
f(x) = x +1 for R
"Verifiy" (sorry my english) that lim x-> 0 (gof)(x) is not equal lim x->0 (gof)(0)
If there is a contradiction with the continuity theorem of composite function, give a justification why.
lim x->0 (gof)(x) = lim x->0 g(x+1) = g(1) = 2
lim x->0(gof)(0) = g(0+1) lim x->0 g(1) = 0
The parts in red are not the way it is usually done.

$\lim _{x \to 0} g \circ f(x) = 2$ BUT $g \circ f(0) = 0\ne 2$ shows that $(gof)$ is not continuous at $x=0~.$

3. ## Re: Composite functions continuity theorem

Hum so gof is not continuous in x=0.

But that contradicts the theorem of continuity?

The only thing i know is that:

if ( $f$is continuous in $a$ and $g$ is continuous in $a$ then $gof$ if continuouse in $a$)

4. ## Re: Composite functions continuity theorem

Originally Posted by Fabio010
Hum so gof is not continuous in x=0.
But that contradicts the theorem of continuity?
But in this case $g$ is not continuous at $x=1$.
$\lim _{x \to 1} g(x) = 2$ but $g(1)=0$.

Originally Posted by Fabio010
if ( $f$is continuous in $a$ and $g$ is continuous in $a$ then $gof$ if continuouse in $a$)
That should be $g$ is continuous at $f(a)$.