Question

@nincompoop

Answer 1

@nincompoop ?

Answer 2

If h(x)=f(x)∙g(x) is continuous at x=0, then f(x) and g(x) are both continuous at x=0. Prove or give a counter example.

Answer 3

In order to contradict this statement, the counterexample must not only fail to satisfy the conclusion that f and g be continuous at 0, but also satisfy the hypothesis that h be continuous at 0; hence h must be defined at 0, and so therefore does g. One may define g by cases:

\( f(x)=0 \) and \(g(x)=\frac{1}{x}\) if \(x \neq \)0 and \(g(x)=0\) if\( x=0\). Then \(h=0\), so h is continuous at 0 but g is not since \( \lim_{x \to 0} g(x) \)does not exist.

\(f(x)=x^2\) and \(g(x)=\frac{1}{x}\) if \(x \neq 0\) and \(g(x)=0\) if \(x=0\). Then h=0, so h is continuous at 0 but g is not since \( \lim_{x \to 0} g(x)\) does not exist.

Answer 4

this is a little clearer

Answer 5

f(x) and g(x) both rely on x, which is x = 0

if what he is saying that only 1 of the functions needs to be continuous to satisfy continuity at f(g(x)) then TRUE

I can give actual examples from textbooks for this to be the case

but this is what I said last night
"if f(x) is continuous at x=a and g(x) is continuous at x=a then the composite function f(g(x)) is continuous at x=a," which is TRUE as well

I did not point out that both functions has to be continuous at x=0 in order to satisfy continuity at x=0 for f(g(x))

Answer 6

I cant remember what u said last night but u summarized it correctly

Answer 7

Basically the statement was false and ya had to give a counter example which I messed up :D

Answer 8

LAUGHING OUT LOUD
you didn't mess up... you just didn't think it through LAUGHING MY ARSE OFF before giving an answer :P