Requirements for continuity of $fleft(frac{1}{g(x)}right)$
$begingroup$
Suppose $f$ and $g$ are continuous images $mathbb{R}rightarrow mathbb{R}$, then the function $fleft(frac{1}{g(x)}right)$ is necesarilly:
- continuous for all $xin mathbb{R}$ for which $xneq0$
- continuous for all $xin mathbb{R}$ for which $f(x)neq0$
- continuous for all $xin mathbb{R}$ for which $g(x)neq0$
- continuous for all $xin mathbb{R}$ for which $frac{1}{g(x)}neq0$
- continuous for all $xin mathbb{R}$ for which $fleft(frac{1}{g(x)}right)neq0$
So apperently there are three possible combination:
- 3
- 3 and 4
- 3, 4 and 5
I understand why 3 has to be true, but not so much why 4 and 5 do not have to be true but that if they are true it is still a correct answer (note these requirements are necessary) so also why can you hav multiple sets of necessary requerments but some sets doe not include other requerments. Does anyone have any idea?
real-analysis continuity
asked Jan 8 at 14:50
ViktorViktor
926
$endgroup$
add a comment |
$begingroup$
Suppose $f$ and $g$ are continuous images $mathbb{R}rightarrow mathbb{R}$, then the function $fleft(frac{1}{g(x)}right)$ is necesarilly:
- continuous for all $xin mathbb{R}$ for which $xneq0$
- continuous for all $xin mathbb{R}$ for which $f(x)neq0$
- continuous for all $xin mathbb{R}$ for which $g(x)neq0$
- continuous for all $xin mathbb{R}$ for which $frac{1}{g(x)}neq0$
- continuous for all $xin mathbb{R}$ for which $fleft(frac{1}{g(x)}right)neq0$
So apperently there are three possible combination:
- 3
- 3 and 4
- 3, 4 and 5
I understand why 3 has to be true, but not so much why 4 and 5 do not have to be true but that if they are true it is still a correct answer (note these requirements are necessary) so also why can you hav multiple sets of necessary requerments but some sets doe not include other requerments. Does anyone have any idea?
real-analysis continuity
asked Jan 8 at 14:50
ViktorViktor
926
$endgroup$
$begingroup$
1,2,4, and 5 simply have nothing to do with the question: continuity doesn't treat $0$ any differently than any other number, so they're just surplus to requirements: the result is still true regardless of being them, but adding them doesn't cause any problems, other than restricting where you can apply your result unnecessarily.
$endgroup$
– user3482749
Jan 8 at 15:01
add a comment |
$begingroup$
Suppose $f$ and $g$ are continuous images $mathbb{R}rightarrow mathbb{R}$, then the function $fleft(frac{1}{g(x)}right)$ is necesarilly:
- continuous for all $xin mathbb{R}$ for which $xneq0$
- continuous for all $xin mathbb{R}$ for which $f(x)neq0$
- continuous for all $xin mathbb{R}$ for which $g(x)neq0$
- continuous for all $xin mathbb{R}$ for which $frac{1}{g(x)}neq0$
- continuous for all $xin mathbb{R}$ for which $fleft(frac{1}{g(x)}right)neq0$
So apperently there are three possible combination:
- 3
- 3 and 4
- 3, 4 and 5
I understand why 3 has to be true, but not so much why 4 and 5 do not have to be true but that if they are true it is still a correct answer (note these requirements are necessary) so also why can you hav multiple sets of necessary requerments but some sets doe not include other requerments. Does anyone have any idea?
real-analysis continuity
asked Jan 8 at 14:50
ViktorViktor
926
$endgroup$
Suppose $f$ and $g$ are continuous images $mathbb{R}rightarrow mathbb{R}$, then the function $fleft(frac{1}{g(x)}right)$ is necesarilly:
- continuous for all $xin mathbb{R}$ for which $xneq0$
- continuous for all $xin mathbb{R}$ for which $f(x)neq0$
- continuous for all $xin mathbb{R}$ for which $g(x)neq0$
- continuous for all $xin mathbb{R}$ for which $frac{1}{g(x)}neq0$
- continuous for all $xin mathbb{R}$ for which $fleft(frac{1}{g(x)}right)neq0$
So apperently there are three possible combination:
- 3
- 3 and 4
- 3, 4 and 5
I understand why 3 has to be true, but not so much why 4 and 5 do not have to be true but that if they are true it is still a correct answer (note these requirements are necessary) so also why can you hav multiple sets of necessary requerments but some sets doe not include other requerments. Does anyone have any idea?
real-analysis continuity
real-analysis continuity
asked Jan 8 at 14:50
ViktorViktor
926
asked Jan 8 at 14:50
ViktorViktor
926
asked Jan 8 at 14:50
ViktorViktor
926
asked Jan 8 at 14:50
ViktorViktor
926
asked Jan 8 at 14:50
ViktorViktor
926
926
$begingroup$
1,2,4, and 5 simply have nothing to do with the question: continuity doesn't treat $0$ any differently than any other number, so they're just surplus to requirements: the result is still true regardless of being them, but adding them doesn't cause any problems, other than restricting where you can apply your result unnecessarily.
$endgroup$
– user3482749
Jan 8 at 15:01
add a comment |
$begingroup$
1,2,4, and 5 simply have nothing to do with the question: continuity doesn't treat $0$ any differently than any other number, so they're just surplus to requirements: the result is still true regardless of being them, but adding them doesn't cause any problems, other than restricting where you can apply your result unnecessarily.
$endgroup$
– user3482749
Jan 8 at 15:01
$begingroup$
1,2,4, and 5 simply have nothing to do with the question: continuity doesn't treat $0$ any differently than any other number, so they're just surplus to requirements: the result is still true regardless of being them, but adding them doesn't cause any problems, other than restricting where you can apply your result unnecessarily.
$endgroup$
– user3482749
Jan 8 at 15:01
$begingroup$
1,2,4, and 5 simply have nothing to do with the question: continuity doesn't treat $0$ any differently than any other number, so they're just surplus to requirements: the result is still true regardless of being them, but adding them doesn't cause any problems, other than restricting where you can apply your result unnecessarily.
$endgroup$
– user3482749
Jan 8 at 15:01
add a comment |
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$begingroup$
1,2,4, and 5 simply have nothing to do with the question: continuity doesn't treat $0$ any differently than any other number, so they're just surplus to requirements: the result is still true regardless of being them, but adding them doesn't cause any problems, other than restricting where you can apply your result unnecessarily.
$endgroup$
– user3482749
Jan 8 at 15:01