#is this proof valid or am i stupid
tacit timberBOT
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lethal knot
$\lim_{x\to 0}(fg)(x)=\infty$ since it's $\frac{1}{x^2}$ for $x\neq 0$
light geodeBOT
Omegabet_
lethal knot
I mean, $f(x)=\begin{cases}7&x\neq 0\1&x=0\end{cases}$ isnt continuous at $0$
light geodeBOT
Omegabet_
lethal knot
you can easily find a g such that (fg)(x)=1 for all x in R and isnt continuous at 1
karmic cedar
you can easily find a g such that (fg)(x)=1 for all x in R and isnt continuous a...
so if i keep the same g
and the new f
what would be the fg
im not sure how to multiply piecewise functions
lethal knot
Clearly it'd be 7/x for x not 0, and 0 for x=0
You do it pointwise
For x!=0, (fg)(x)=7 * 1/x
For x =0, (fg)(0)=1*0=0
lethal knot
Yep
karmic cedar
so the claim is true?
lethal knot
No
karmic cedar
but to proof the claim being wrong i need to show its continuous
karmic cedar
oh
lethal knot
No
karmic cedar
i have to change g(X)
lethal knot
Yes...
karmic cedar
ok so if i just make g(X) the opposite
lethal knot
Explicitly write g
karmic cedar
so for x=0 7 and wiseversa
lethal knot
Opposite isn't well defined
lethal knot
Opposite doesn't have a meaning here
karmic cedar
g(X) 1 x=/=0
and 7 =0
then H(X) would be 7 everywhere
and therefore be continuous
lethal knot
Yes
karmic cedar
ok thanks alot
gentle tuskBOT
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light geodeBOT
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cerulean wadi
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