Circular reasoning in L'Hopital's rule The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)L'Hospital's Rule Question.Why doesn't L'Hopital's rule work in this case?Finding limits by L'Hospital's RuleApplication of l'hospital rule to exponential functionL'Hopital's Rule helpAvoiding circular logic using L'Hospital's ruleShow $ lim_t to infty -t^xe^-t = 0 $Applying L'Hopital's Rule without hypothesis on numeratorHow to show that $lim_xto infty left(int^x_2 (ln t)^-1 dt right) big/ (x /ln x)=1$?$lim_x to infty e^x - frace^xx+1$ Application of L'Hopital's Rule
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Circular reasoning in L'Hopital's rule
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)L'Hospital's Rule Question.Why doesn't L'Hopital's rule work in this case?Finding limits by L'Hospital's RuleApplication of l'hospital rule to exponential functionL'Hopital's Rule helpAvoiding circular logic using L'Hospital's ruleShow $ lim_t to infty -t^xe^-t = 0 $Applying L'Hopital's Rule without hypothesis on numeratorHow to show that $lim_xto infty left(int^x_2 (ln t)^-1 dt right) big/ (x /ln x)=1$?$lim_x to infty e^x - frace^xx+1$ Application of L'Hopital's Rule
$begingroup$
Suppose we have a function $f(x)$ that satisfies:
$$lim_xtoinftyf(x)=L$$
Where $LinmathbbR$. Is this true?
$$lim_xtoinftyf'(x)=0$$
My approach was simply this:
$$lim_xtoinftyf(x)=lim_xtoinftyfracxf(x)x=L$$
And applying L'Hospital's rule we have:
$$lim_xtoinftyfracxf(x)x=lim_xtoinftyfracf(x)+xf'(x)1=L$$
$$lim_xtoinftyf(x)+xf'(x)=L+lim_xtoinftyxf'(x)=L$$
And finally:
$$lim_xtoinftyxf'(x)=0$$
Now, the only way this is possible is if $lim_xtoinftyf'(x)neqinfty$ and $lim_xtoinftyf'(x)neq AinmathbbR$ , because otherways the $lim_xtoinftyxf'(x)$ would go to infinity. In conclusion, $lim_xtoinftyf'(x)=0$
Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.
limits
asked yesterday
marcozzmarcozz
137110
$endgroup$
add a comment |
$begingroup$
Suppose we have a function $f(x)$ that satisfies:
$$lim_xtoinftyf(x)=L$$
Where $LinmathbbR$. Is this true?
$$lim_xtoinftyf'(x)=0$$
My approach was simply this:
$$lim_xtoinftyf(x)=lim_xtoinftyfracxf(x)x=L$$
And applying L'Hospital's rule we have:
$$lim_xtoinftyfracxf(x)x=lim_xtoinftyfracf(x)+xf'(x)1=L$$
$$lim_xtoinftyf(x)+xf'(x)=L+lim_xtoinftyxf'(x)=L$$
And finally:
$$lim_xtoinftyxf'(x)=0$$
Now, the only way this is possible is if $lim_xtoinftyf'(x)neqinfty$ and $lim_xtoinftyf'(x)neq AinmathbbR$ , because otherways the $lim_xtoinftyxf'(x)$ would go to infinity. In conclusion, $lim_xtoinftyf'(x)=0$
Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.
limits
asked yesterday
marcozzmarcozz
137110
$endgroup$
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
yesterday
$begingroup$
To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=sin x^2$ has no $xtoinfty$ limit. If $L$ were nonzero, on the other hand...
$endgroup$
– J.G.
yesterday
2
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_xtoinfty frac2-1/x1-1/x$.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
yesterday
add a comment |
$begingroup$
Suppose we have a function $f(x)$ that satisfies:
$$lim_xtoinftyf(x)=L$$
Where $LinmathbbR$. Is this true?
$$lim_xtoinftyf'(x)=0$$
My approach was simply this:
$$lim_xtoinftyf(x)=lim_xtoinftyfracxf(x)x=L$$
And applying L'Hospital's rule we have:
$$lim_xtoinftyfracxf(x)x=lim_xtoinftyfracf(x)+xf'(x)1=L$$
$$lim_xtoinftyf(x)+xf'(x)=L+lim_xtoinftyxf'(x)=L$$
And finally:
$$lim_xtoinftyxf'(x)=0$$
Now, the only way this is possible is if $lim_xtoinftyf'(x)neqinfty$ and $lim_xtoinftyf'(x)neq AinmathbbR$ , because otherways the $lim_xtoinftyxf'(x)$ would go to infinity. In conclusion, $lim_xtoinftyf'(x)=0$
Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.
limits
asked yesterday
marcozzmarcozz
137110
$endgroup$
Suppose we have a function $f(x)$ that satisfies:
$$lim_xtoinftyf(x)=L$$
Where $LinmathbbR$. Is this true?
$$lim_xtoinftyf'(x)=0$$
My approach was simply this:
$$lim_xtoinftyf(x)=lim_xtoinftyfracxf(x)x=L$$
And applying L'Hospital's rule we have:
$$lim_xtoinftyfracxf(x)x=lim_xtoinftyfracf(x)+xf'(x)1=L$$
$$lim_xtoinftyf(x)+xf'(x)=L+lim_xtoinftyxf'(x)=L$$
And finally:
$$lim_xtoinftyxf'(x)=0$$
Now, the only way this is possible is if $lim_xtoinftyf'(x)neqinfty$ and $lim_xtoinftyf'(x)neq AinmathbbR$ , because otherways the $lim_xtoinftyxf'(x)$ would go to infinity. In conclusion, $lim_xtoinftyf'(x)=0$
Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.
limits
limits
asked yesterday
marcozzmarcozz
137110
asked yesterday
marcozzmarcozz
137110
edited yesterday
marcozz
asked yesterday
marcozzmarcozz
137110
asked yesterday
marcozzmarcozz
137110
asked yesterday
marcozzmarcozz
137110
137110
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
yesterday
$begingroup$
To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=sin x^2$ has no $xtoinfty$ limit. If $L$ were nonzero, on the other hand...
$endgroup$
– J.G.
yesterday
2
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_xtoinfty frac2-1/x1-1/x$.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
yesterday
add a comment |
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
yesterday
$begingroup$
To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=sin x^2$ has no $xtoinfty$ limit. If $L$ were nonzero, on the other hand...
$endgroup$
– J.G.
yesterday
2
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_xtoinfty frac2-1/x1-1/x$.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
yesterday
3
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
yesterday
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
yesterday
$begingroup$
To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=sin x^2$ has no $xtoinfty$ limit. If $L$ were nonzero, on the other hand...
$endgroup$
– J.G.
yesterday
$begingroup$
To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=sin x^2$ has no $xtoinfty$ limit. If $L$ were nonzero, on the other hand...
$endgroup$
– J.G.
yesterday
2
2
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_xtoinfty frac2-1/x1-1/x$.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_xtoinfty frac2-1/x1-1/x$.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
yesterday
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that $f(x)=dfracsin(x^2)x$. Then $lim_xtoinftyf(x)=0$, but the limit $lim_xtoinftyf'(x)$ doesn't exist.
If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_xtoinftyfracxsin(x^2)x^2.$$But if $g(x)=xsin(x^2)$, then the limit $lim_xtoinftyg'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
10
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
$endgroup$
– user21820
22 hours ago
add a comment |
$begingroup$
(Paraphrased from Wikipedia.)
L'Hôpital's rule:
Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c in I$, if
$$ lim _x to cF(x)=lim _xto cG(x)=0 text or pm infty, tag1. $$
$$ G'(x)neq 0 text for all x in I, text with x ne c, text and tag2. $$
$$ lim_x to cfracF'(x)G'(x) text exists. tag3. $$
then
$$lim_x to c fracF(x)G(x) =lim_x to c fracF'(x)G'(x). tag4.$$
You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.
Since $lim _xto inftyG(x)= infty$, condition $(1.)$ requires that
$$lim _x to inftyxf(x) = infty. tagA.$$
Condition $(2.)$ is satisfied by $G(x)=x$.
Condition $(3.)$ requires that
$$lim_x to infty[f(x)+xf'(x)] text exists. tagB.$$
If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule,
$$ lim_x to infty f(x) = lim_x to infty[f(x)+xf'(x)]$$
Others have shown you that counter examples do exists.
answered yesterday
steven gregorysteven gregory
18.4k32359
$endgroup$
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Suppose that $f(x)=dfracsin(x^2)x$. Then $lim_xtoinftyf(x)=0$, but the limit $lim_xtoinftyf'(x)$ doesn't exist.
If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_xtoinftyfracxsin(x^2)x^2.$$But if $g(x)=xsin(x^2)$, then the limit $lim_xtoinftyg'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
10
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
$endgroup$
– user21820
22 hours ago
add a comment |
$begingroup$
Suppose that $f(x)=dfracsin(x^2)x$. Then $lim_xtoinftyf(x)=0$, but the limit $lim_xtoinftyf'(x)$ doesn't exist.
If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_xtoinftyfracxsin(x^2)x^2.$$But if $g(x)=xsin(x^2)$, then the limit $lim_xtoinftyg'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
10
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
$endgroup$
– user21820
22 hours ago
add a comment |
$begingroup$
Suppose that $f(x)=dfracsin(x^2)x$. Then $lim_xtoinftyf(x)=0$, but the limit $lim_xtoinftyf'(x)$ doesn't exist.
If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_xtoinftyfracxsin(x^2)x^2.$$But if $g(x)=xsin(x^2)$, then the limit $lim_xtoinftyg'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
Suppose that $f(x)=dfracsin(x^2)x$. Then $lim_xtoinftyf(x)=0$, but the limit $lim_xtoinftyf'(x)$ doesn't exist.
If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_xtoinftyfracxsin(x^2)x^2.$$But if $g(x)=xsin(x^2)$, then the limit $lim_xtoinftyg'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
answered yesterday
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
10
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
$endgroup$
– user21820
22 hours ago
add a comment |
10
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
$endgroup$
– user21820
22 hours ago
10
10
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
yesterday
2
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
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– Mark Viola
yesterday
1
1
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@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
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– user21820
22 hours ago
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@marcozz: I must say that it's rather disturbing that you are being taught L'Hopital's rule without also the precise conditions under which it holds. This kind of bad pedagogy is the same reason for the fallacy $1 = sqrt1 = sqrt-1·-1 = sqrt-1·sqrt-1 = i·i = -1$. If all rules were taught properly, such fallacies would never even arise.
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– user21820
22 hours ago
add a comment |
$begingroup$
(Paraphrased from Wikipedia.)
L'Hôpital's rule:
Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c in I$, if
$$ lim _x to cF(x)=lim _xto cG(x)=0 text or pm infty, tag1. $$
$$ G'(x)neq 0 text for all x in I, text with x ne c, text and tag2. $$
$$ lim_x to cfracF'(x)G'(x) text exists. tag3. $$
then
$$lim_x to c fracF(x)G(x) =lim_x to c fracF'(x)G'(x). tag4.$$
You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.
Since $lim _xto inftyG(x)= infty$, condition $(1.)$ requires that
$$lim _x to inftyxf(x) = infty. tagA.$$
Condition $(2.)$ is satisfied by $G(x)=x$.
Condition $(3.)$ requires that
$$lim_x to infty[f(x)+xf'(x)] text exists. tagB.$$
If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule,
$$ lim_x to infty f(x) = lim_x to infty[f(x)+xf'(x)]$$
Others have shown you that counter examples do exists.
answered yesterday
steven gregorysteven gregory
18.4k32359
$endgroup$
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
add a comment |
$begingroup$
(Paraphrased from Wikipedia.)
L'Hôpital's rule:
Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c in I$, if
$$ lim _x to cF(x)=lim _xto cG(x)=0 text or pm infty, tag1. $$
$$ G'(x)neq 0 text for all x in I, text with x ne c, text and tag2. $$
$$ lim_x to cfracF'(x)G'(x) text exists. tag3. $$
then
$$lim_x to c fracF(x)G(x) =lim_x to c fracF'(x)G'(x). tag4.$$
You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.
Since $lim _xto inftyG(x)= infty$, condition $(1.)$ requires that
$$lim _x to inftyxf(x) = infty. tagA.$$
Condition $(2.)$ is satisfied by $G(x)=x$.
Condition $(3.)$ requires that
$$lim_x to infty[f(x)+xf'(x)] text exists. tagB.$$
If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule,
$$ lim_x to infty f(x) = lim_x to infty[f(x)+xf'(x)]$$
Others have shown you that counter examples do exists.
answered yesterday
steven gregorysteven gregory
18.4k32359
$endgroup$
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
add a comment |
$begingroup$
(Paraphrased from Wikipedia.)
L'Hôpital's rule:
Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c in I$, if
$$ lim _x to cF(x)=lim _xto cG(x)=0 text or pm infty, tag1. $$
$$ G'(x)neq 0 text for all x in I, text with x ne c, text and tag2. $$
$$ lim_x to cfracF'(x)G'(x) text exists. tag3. $$
then
$$lim_x to c fracF(x)G(x) =lim_x to c fracF'(x)G'(x). tag4.$$
You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.
Since $lim _xto inftyG(x)= infty$, condition $(1.)$ requires that
$$lim _x to inftyxf(x) = infty. tagA.$$
Condition $(2.)$ is satisfied by $G(x)=x$.
Condition $(3.)$ requires that
$$lim_x to infty[f(x)+xf'(x)] text exists. tagB.$$
If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule,
$$ lim_x to infty f(x) = lim_x to infty[f(x)+xf'(x)]$$
Others have shown you that counter examples do exists.
answered yesterday
steven gregorysteven gregory
18.4k32359
$endgroup$
(Paraphrased from Wikipedia.)
L'Hôpital's rule:
Given functions $f$ and $g$ which are differentiable on an open interval $I$, except possibly at a point $c in I$, if
$$ lim _x to cF(x)=lim _xto cG(x)=0 text or pm infty, tag1. $$
$$ G'(x)neq 0 text for all x in I, text with x ne c, text and tag2. $$
$$ lim_x to cfracF'(x)G'(x) text exists. tag3. $$
then
$$lim_x to c fracF(x)G(x) =lim_x to c fracF'(x)G'(x). tag4.$$
You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.
Since $lim _xto inftyG(x)= infty$, condition $(1.)$ requires that
$$lim _x to inftyxf(x) = infty. tagA.$$
Condition $(2.)$ is satisfied by $G(x)=x$.
Condition $(3.)$ requires that
$$lim_x to infty[f(x)+xf'(x)] text exists. tagB.$$
If conditions $(A.)$ and $(B.)$ are met, then, by L'Hôpital's rule,
$$ lim_x to infty f(x) = lim_x to infty[f(x)+xf'(x)]$$
Others have shown you that counter examples do exists.
answered yesterday
steven gregorysteven gregory
18.4k32359
answered yesterday
steven gregorysteven gregory
18.4k32359
answered yesterday
steven gregorysteven gregory
18.4k32359
answered yesterday
steven gregorysteven gregory
18.4k32359
18.4k32359
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
add a comment |
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
1
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
$begingroup$
Application of LHR does not require that the numerator approach $infty$. In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR. This answer makes that fact explicit.
$endgroup$
– Mark Viola
yesterday
1
1
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limitsx to infty G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
$begingroup$
First, I don't presume that the OP is a male. My point is that in "Condition $1$ can be relaxed. It is NOT required that $lim F=infty$.
$endgroup$
– Mark Viola
yesterday
add a comment |
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-limits
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
yesterday
$begingroup$
To use the rule you'd need $xf$, like $x$, to diverge; but in the counterexample others have discussed herein, $xf=sin x^2$ has no $xtoinfty$ limit. If $L$ were nonzero, on the other hand...
$endgroup$
– J.G.
yesterday
2
$begingroup$
Application of LHR does not require that the numerator approach ∞ ∞ . In fact, the limit of the numerator need not even exist. What IS required is that the limit of the quotient of derivatives DOES exist. In the counter examples given on this page, that limit fails to exist and therefore invalidates application of LHR.
$endgroup$
– Mark Viola
yesterday
$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_xtoinfty frac2-1/x1-1/x$.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
yesterday