Limit of the composistion of two functions when the limit of $f$ does not exist?
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The theorem about the limit of composition of two real functions $f$ and $g$ is proved here. But it is required that the two limits (of $f$ and $g$) both exist.
I can't understand how to deal with the case in which the limit of $f$ does not exists. In particular I would like to know if the following is correct.
Consider $f(x)$ and $g(x)$ (real functions).
If I find out that $lim_{x to x_0} f(x)$ does not exists, can I conclude that $lim_{xto x_0 }g(f(x))$ does not exist?
Under what conditions is this correct?
calculus limits function-and-relation-composition
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
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add a comment |
$begingroup$
The theorem about the limit of composition of two real functions $f$ and $g$ is proved here. But it is required that the two limits (of $f$ and $g$) both exist.
I can't understand how to deal with the case in which the limit of $f$ does not exists. In particular I would like to know if the following is correct.
Consider $f(x)$ and $g(x)$ (real functions).
If I find out that $lim_{x to x_0} f(x)$ does not exists, can I conclude that $lim_{xto x_0 }g(f(x))$ does not exist?
Under what conditions is this correct?
calculus limits function-and-relation-composition
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
$endgroup$
$begingroup$
See the theorem at the end of the answer math.stackexchange.com/a/1073047/72031
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– Paramanand Singh
Dec 9 '18 at 5:10
add a comment |
$begingroup$
The theorem about the limit of composition of two real functions $f$ and $g$ is proved here. But it is required that the two limits (of $f$ and $g$) both exist.
I can't understand how to deal with the case in which the limit of $f$ does not exists. In particular I would like to know if the following is correct.
Consider $f(x)$ and $g(x)$ (real functions).
If I find out that $lim_{x to x_0} f(x)$ does not exists, can I conclude that $lim_{xto x_0 }g(f(x))$ does not exist?
Under what conditions is this correct?
calculus limits function-and-relation-composition
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
$endgroup$
The theorem about the limit of composition of two real functions $f$ and $g$ is proved here. But it is required that the two limits (of $f$ and $g$) both exist.
I can't understand how to deal with the case in which the limit of $f$ does not exists. In particular I would like to know if the following is correct.
Consider $f(x)$ and $g(x)$ (real functions).
If I find out that $lim_{x to x_0} f(x)$ does not exists, can I conclude that $lim_{xto x_0 }g(f(x))$ does not exist?
Under what conditions is this correct?
calculus limits function-and-relation-composition
calculus limits function-and-relation-composition
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
asked Dec 8 '18 at 18:33
GianolepoGianolepo
775918
775918
$begingroup$
See the theorem at the end of the answer math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Dec 9 '18 at 5:10
add a comment |
$begingroup$
See the theorem at the end of the answer math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Dec 9 '18 at 5:10
$begingroup$
See the theorem at the end of the answer math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Dec 9 '18 at 5:10
$begingroup$
See the theorem at the end of the answer math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Dec 9 '18 at 5:10
add a comment |
1 Answer
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No we can't, let consider as $x to infty$
- $f(x)=sin x$
- $g(x)=1$
but for $g(x)=x$ of course the implication holds.
More in general, I think we need to consider case by case upon the specific functions we are considering. I'm not aware about and I can't exclude the existence of some strange pathological cases.
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
No we can't, let consider as $x to infty$
- $f(x)=sin x$
- $g(x)=1$
but for $g(x)=x$ of course the implication holds.
More in general, I think we need to consider case by case upon the specific functions we are considering. I'm not aware about and I can't exclude the existence of some strange pathological cases.
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
No we can't, let consider as $x to infty$
- $f(x)=sin x$
- $g(x)=1$
but for $g(x)=x$ of course the implication holds.
More in general, I think we need to consider case by case upon the specific functions we are considering. I'm not aware about and I can't exclude the existence of some strange pathological cases.
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
No we can't, let consider as $x to infty$
- $f(x)=sin x$
- $g(x)=1$
but for $g(x)=x$ of course the implication holds.
More in general, I think we need to consider case by case upon the specific functions we are considering. I'm not aware about and I can't exclude the existence of some strange pathological cases.
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
$endgroup$
No we can't, let consider as $x to infty$
- $f(x)=sin x$
- $g(x)=1$
but for $g(x)=x$ of course the implication holds.
More in general, I think we need to consider case by case upon the specific functions we are considering. I'm not aware about and I can't exclude the existence of some strange pathological cases.
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
edited Dec 8 '18 at 18:42
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
answered Dec 8 '18 at 18:35
gimusigimusi
92.9k94494
92.9k94494
add a comment |
add a comment |
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See the theorem at the end of the answer math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Dec 9 '18 at 5:10