Circular reasoning in L'Hopital's rule 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
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 Apr 12 at 13:26
marcozzmarcozz
137111
$endgroup$
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 Apr 12 at 13:26
marcozzmarcozz
137111
$endgroup$
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
Apr 12 at 13:33
$begingroup$
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.
Apr 12 at 15:24
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
Apr 12 at 16:45
$begingroup$
@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
Apr 12 at 20:50
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
Apr 12 at 22:03
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 Apr 12 at 13:26
marcozzmarcozz
137111
$endgroup$
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 Apr 12 at 13:26
marcozzmarcozz
137111
asked Apr 12 at 13:26
marcozzmarcozz
137111
edited Apr 12 at 16:59
marcozz
asked Apr 12 at 13:26
marcozzmarcozz
137111
asked Apr 12 at 13:26
marcozzmarcozz
137111
asked Apr 12 at 13:26
marcozzmarcozz
137111
137111
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
Apr 12 at 13:33
$begingroup$
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.
Apr 12 at 15:24
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
Apr 12 at 16:45
$begingroup$
@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
Apr 12 at 20:50
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
Apr 12 at 22:03
add a comment |
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
Apr 12 at 13:33
$begingroup$
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.
Apr 12 at 15:24
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
Apr 12 at 16:45
$begingroup$
@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
Apr 12 at 20:50
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
Apr 12 at 22:03
3
3
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
Apr 12 at 13:33
$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
Apr 12 at 13:33
$begingroup$
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.
Apr 12 at 15:24
$begingroup$
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.
Apr 12 at 15:24
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
Apr 12 at 16:45
$begingroup$
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
Apr 12 at 16:45
$begingroup$
@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
Apr 12 at 20:50
$begingroup$
@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
Apr 12 at 20:50
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
Apr 12 at 22:03
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
Apr 12 at 22:03
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 Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
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
Apr 12 at 13:39
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
Apr 12 at 14:10
$begingroup$
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
Apr 12 at 16:45
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
Apr 13 at 3:09
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 Apr 12 at 14:53
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
Apr 12 at 16:45
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
Apr 12 at 21:23
$begingroup$
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
Apr 12 at 22:08
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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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 Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
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
Apr 12 at 13:39
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
Apr 12 at 14:10
$begingroup$
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
Apr 12 at 16:45
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
Apr 13 at 3:09
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 Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
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
Apr 12 at 13:39
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
Apr 12 at 14:10
$begingroup$
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
Apr 12 at 16:45
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
Apr 13 at 3:09
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 Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
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 Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 12 at 13:33
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
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
Apr 12 at 13:39
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
Apr 12 at 14:10
$begingroup$
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
Apr 12 at 16:45
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
Apr 13 at 3:09
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
Apr 12 at 13:39
2
$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
Apr 12 at 14:10
$begingroup$
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
Apr 12 at 16:45
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
Apr 13 at 3:09
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
Apr 12 at 13:39
$begingroup$
so the only thing we can conclude is: if $lim_x to infty f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
Apr 12 at 13:39
2
2
$begingroup$
@antkam Yes, that is correct.
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– José Carlos Santos
Apr 12 at 14:10
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@antkam Yes, that is correct.
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– José Carlos Santos
Apr 12 at 14:10
$begingroup$
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
Apr 12 at 16:45
$begingroup$
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
Apr 12 at 16:45
1
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.
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– user21820
Apr 13 at 3:09
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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.
$endgroup$
– user21820
Apr 13 at 3:09
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 Apr 12 at 14:53
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
Apr 12 at 16:45
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
Apr 12 at 21:23
$begingroup$
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
Apr 12 at 22:08
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 Apr 12 at 14:53
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
Apr 12 at 16:45
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
Apr 12 at 21:23
$begingroup$
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
Apr 12 at 22:08
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 Apr 12 at 14:53
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 Apr 12 at 14:53
steven gregorysteven gregory
18.4k32359
answered Apr 12 at 14:53
steven gregorysteven gregory
18.4k32359
answered Apr 12 at 14:53
steven gregorysteven gregory
18.4k32359
answered Apr 12 at 14:53
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
Apr 12 at 16:45
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
Apr 12 at 21:23
$begingroup$
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
Apr 12 at 22:08
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
Apr 12 at 16:45
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
Apr 12 at 21:23
$begingroup$
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
Apr 12 at 22:08
$begingroup$
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
Apr 12 at 16:45
$begingroup$
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
Apr 12 at 16:45
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
Apr 12 at 21:23
$begingroup$
@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
Apr 12 at 21:23
$begingroup$
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
Apr 12 at 22:08
$begingroup$
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
Apr 12 at 22:08
add a comment |
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$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
Apr 12 at 13:33
$begingroup$
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.
Apr 12 at 15:24
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
Apr 12 at 16:45
$begingroup$
@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
Apr 12 at 20:50
$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
Apr 12 at 22:03