Circular reasoning in L'Hopital's rule












7












$begingroup$


Suppose we have a function $f(x)$ that satisfies:
$$lim_{xtoinfty}f(x)=L$$
Where $Linmathbb{R}$. Is this true?
$$lim_{xtoinfty}f'(x)=0$$



My approach was simply this:



$$lim_{xtoinfty}f(x)=lim_{xtoinfty}frac{xf(x)}{x}=L$$



And applying L'Hospital's rule we have:



$$lim_{xtoinfty}frac{xf(x)}{x}=lim_{xtoinfty}frac{f(x)+xf'(x)}{1}=L$$
$$lim_{xtoinfty}f(x)+xf'(x)=L+lim_{xtoinfty}xf'(x)=L$$
And finally:
$$lim_{xtoinfty}xf'(x)=0$$
Now, the only way this is possible is if $lim_{xtoinfty}f'(x)neqinfty$ and $lim_{xtoinfty}f'(x)neq Ainmathbb{R}$ , because otherways the $lim_{xtoinfty}xf'(x)$ would go to infinity. In conclusion, $lim_{xtoinfty}f'(x)=0$



Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.






limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
    $endgroup$
    – John Doe
    13 hours ago












  • $begingroup$
    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.
    12 hours ago






  • 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
    10 hours ago










  • $begingroup$
    @MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_{xtoinfty} frac{2-1/x}{1-1/x}$.
    $endgroup$
    – Henning Makholm
    6 hours ago












  • $begingroup$
    @HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
    $endgroup$
    – Mark Viola
    5 hours ago
















7












$begingroup$


Suppose we have a function $f(x)$ that satisfies:
$$lim_{xtoinfty}f(x)=L$$
Where $Linmathbb{R}$. Is this true?
$$lim_{xtoinfty}f'(x)=0$$



My approach was simply this:



$$lim_{xtoinfty}f(x)=lim_{xtoinfty}frac{xf(x)}{x}=L$$



And applying L'Hospital's rule we have:



$$lim_{xtoinfty}frac{xf(x)}{x}=lim_{xtoinfty}frac{f(x)+xf'(x)}{1}=L$$
$$lim_{xtoinfty}f(x)+xf'(x)=L+lim_{xtoinfty}xf'(x)=L$$
And finally:
$$lim_{xtoinfty}xf'(x)=0$$
Now, the only way this is possible is if $lim_{xtoinfty}f'(x)neqinfty$ and $lim_{xtoinfty}f'(x)neq Ainmathbb{R}$ , because otherways the $lim_{xtoinfty}xf'(x)$ would go to infinity. In conclusion, $lim_{xtoinfty}f'(x)=0$



Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.






limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
    $endgroup$
    – John Doe
    13 hours ago












  • $begingroup$
    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.
    12 hours ago






  • 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
    10 hours ago










  • $begingroup$
    @MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_{xtoinfty} frac{2-1/x}{1-1/x}$.
    $endgroup$
    – Henning Makholm
    6 hours ago












  • $begingroup$
    @HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
    $endgroup$
    – Mark Viola
    5 hours ago














7












7








7


4



$begingroup$


Suppose we have a function $f(x)$ that satisfies:
$$lim_{xtoinfty}f(x)=L$$
Where $Linmathbb{R}$. Is this true?
$$lim_{xtoinfty}f'(x)=0$$



My approach was simply this:



$$lim_{xtoinfty}f(x)=lim_{xtoinfty}frac{xf(x)}{x}=L$$



And applying L'Hospital's rule we have:



$$lim_{xtoinfty}frac{xf(x)}{x}=lim_{xtoinfty}frac{f(x)+xf'(x)}{1}=L$$
$$lim_{xtoinfty}f(x)+xf'(x)=L+lim_{xtoinfty}xf'(x)=L$$
And finally:
$$lim_{xtoinfty}xf'(x)=0$$
Now, the only way this is possible is if $lim_{xtoinfty}f'(x)neqinfty$ and $lim_{xtoinfty}f'(x)neq Ainmathbb{R}$ , because otherways the $lim_{xtoinfty}xf'(x)$ would go to infinity. In conclusion, $lim_{xtoinfty}f'(x)=0$



Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.






limits







share|cite|improve this question











$endgroup$




Suppose we have a function $f(x)$ that satisfies:
$$lim_{xtoinfty}f(x)=L$$
Where $Linmathbb{R}$. Is this true?
$$lim_{xtoinfty}f'(x)=0$$



My approach was simply this:



$$lim_{xtoinfty}f(x)=lim_{xtoinfty}frac{xf(x)}{x}=L$$



And applying L'Hospital's rule we have:



$$lim_{xtoinfty}frac{xf(x)}{x}=lim_{xtoinfty}frac{f(x)+xf'(x)}{1}=L$$
$$lim_{xtoinfty}f(x)+xf'(x)=L+lim_{xtoinfty}xf'(x)=L$$
And finally:
$$lim_{xtoinfty}xf'(x)=0$$
Now, the only way this is possible is if $lim_{xtoinfty}f'(x)neqinfty$ and $lim_{xtoinfty}f'(x)neq Ainmathbb{R}$ , because otherways the $lim_{xtoinfty}xf'(x)$ would go to infinity. In conclusion, $lim_{xtoinfty}f'(x)=0$



Is this in any way circular reasoning? I'm especially worried about the part when we apply the L'Hospital's rule.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago







marcozz

















asked 13 hours ago









marcozzmarcozz

135110




135110








  • 3




    $begingroup$
    Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
    $endgroup$
    – John Doe
    13 hours ago












  • $begingroup$
    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.
    12 hours ago






  • 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
    10 hours ago










  • $begingroup$
    @MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_{xtoinfty} frac{2-1/x}{1-1/x}$.
    $endgroup$
    – Henning Makholm
    6 hours ago












  • $begingroup$
    @HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
    $endgroup$
    – Mark Viola
    5 hours ago














  • 3




    $begingroup$
    Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
    $endgroup$
    – John Doe
    13 hours ago












  • $begingroup$
    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.
    12 hours ago






  • 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
    10 hours ago










  • $begingroup$
    @MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_{xtoinfty} frac{2-1/x}{1-1/x}$.
    $endgroup$
    – Henning Makholm
    6 hours ago












  • $begingroup$
    @HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
    $endgroup$
    – Mark Viola
    5 hours ago








3




3




$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
13 hours ago






$begingroup$
Something like $f(x)=sin(x^2)/x$ provides a counterexample, doesn't it?
$endgroup$
– John Doe
13 hours ago














$begingroup$
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.
12 hours ago




$begingroup$
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.
12 hours ago




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
10 hours ago




$begingroup$
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
10 hours ago












$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_{xtoinfty} frac{2-1/x}{1-1/x}$.
$endgroup$
– Henning Makholm
6 hours ago






$begingroup$
@MarkViola: On the other hand, then you do need the denominator to approach $infty$. Otherwise you get into trouble with cases like $limlimits_{xtoinfty} frac{2-1/x}{1-1/x}$.
$endgroup$
– Henning Makholm
6 hours ago














$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
5 hours ago




$begingroup$
@HenningMakholm Indeed. The limit of the denominator must approach $infyt$ (or $-infty$).
$endgroup$
– Mark Viola
5 hours ago










2 Answers
2






active

oldest

votes


















15












$begingroup$

Suppose that $f(x)=dfrac{sin(x^2)}x$. Then $lim_{xtoinfty}f(x)=0$, but the limit $lim_{xtoinfty}f'(x)$ doesn't exist.



If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_{xtoinfty}frac{xsin(x^2)}{x^2}.$$But if $g(x)=xsin(x^2)$, then the limit $lim_{xtoinfty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.






share|cite|improve this answer









$endgroup$









  • 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
    13 hours ago












  • $begingroup$
    Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
    $endgroup$
    – marcozz
    13 hours ago








  • 2




    $begingroup$
    @antkam Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    @marcozz Indeed.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    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
    10 hours ago



















3












$begingroup$

(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



$$ lim _{x to c}F(x)=lim _{xto c}G(x)=0 text{ or }pm infty, tag{1.} $$
$$ G'(x)neq 0 text{ for all }x in I, text{ with }x ne c, text{ and} tag{2.} $$
$$ lim_{x to c}frac{F'(x)}{G'(x)} text{ exists.} tag{3.} $$



then



$$lim_{x to c} frac{F(x)}{G(x)} =lim_{x to c} frac{F'(x)}{G'(x)}. tag{4.}$$



You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.



Since $lim _{xto infty}G(x)= infty$, condition $(1.)$ requires that
$$lim _{x to infty}xf(x) = infty. tag{A.}$$



Condition $(2.)$ is satisfied by $G(x)=x$.



Condition $(3.)$ requires that
$$lim_{x to infty}[f(x)+xf'(x)] text{ exists.} tag{B.}$$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    10 hours ago












  • $begingroup$
    @MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
    $endgroup$
    – steven gregory
    6 hours ago












  • $begingroup$
    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
    5 hours ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









15












$begingroup$

Suppose that $f(x)=dfrac{sin(x^2)}x$. Then $lim_{xtoinfty}f(x)=0$, but the limit $lim_{xtoinfty}f'(x)$ doesn't exist.



If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_{xtoinfty}frac{xsin(x^2)}{x^2}.$$But if $g(x)=xsin(x^2)$, then the limit $lim_{xtoinfty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.






share|cite|improve this answer









$endgroup$









  • 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
    13 hours ago












  • $begingroup$
    Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
    $endgroup$
    – marcozz
    13 hours ago








  • 2




    $begingroup$
    @antkam Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    @marcozz Indeed.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    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
    10 hours ago
















15












$begingroup$

Suppose that $f(x)=dfrac{sin(x^2)}x$. Then $lim_{xtoinfty}f(x)=0$, but the limit $lim_{xtoinfty}f'(x)$ doesn't exist.



If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_{xtoinfty}frac{xsin(x^2)}{x^2}.$$But if $g(x)=xsin(x^2)$, then the limit $lim_{xtoinfty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.






share|cite|improve this answer









$endgroup$









  • 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
    13 hours ago












  • $begingroup$
    Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
    $endgroup$
    – marcozz
    13 hours ago








  • 2




    $begingroup$
    @antkam Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    @marcozz Indeed.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    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
    10 hours ago














15












15








15





$begingroup$

Suppose that $f(x)=dfrac{sin(x^2)}x$. Then $lim_{xtoinfty}f(x)=0$, but the limit $lim_{xtoinfty}f'(x)$ doesn't exist.



If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_{xtoinfty}frac{xsin(x^2)}{x^2}.$$But if $g(x)=xsin(x^2)$, then the limit $lim_{xtoinfty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.






share|cite|improve this answer









$endgroup$



Suppose that $f(x)=dfrac{sin(x^2)}x$. Then $lim_{xtoinfty}f(x)=0$, but the limit $lim_{xtoinfty}f'(x)$ doesn't exist.



If you try to apply L'Hopital's Rule here as you did, you will be working with$$lim_{xtoinfty}frac{xsin(x^2)}{x^2}.$$But if $g(x)=xsin(x^2)$, then the limit $lim_{xtoinfty}g'(x)$ doesn't exist. Therefore, you cannot apply L'Hopital's Rule here.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 13 hours ago









José Carlos SantosJosé Carlos Santos

174k23133243




174k23133243








  • 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
    13 hours ago












  • $begingroup$
    Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
    $endgroup$
    – marcozz
    13 hours ago








  • 2




    $begingroup$
    @antkam Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    @marcozz Indeed.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    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
    10 hours ago














  • 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
    13 hours ago












  • $begingroup$
    Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
    $endgroup$
    – marcozz
    13 hours ago








  • 2




    $begingroup$
    @antkam Yes, that is correct.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    @marcozz Indeed.
    $endgroup$
    – José Carlos Santos
    13 hours ago










  • $begingroup$
    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
    10 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
13 hours ago






$begingroup$
so the only thing we can conclude is: if $lim_{x to infty} f'(x)$ exists, then it must be $0$?
$endgroup$
– antkam
13 hours ago














$begingroup$
Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
$endgroup$
– marcozz
13 hours ago






$begingroup$
Thanks for the counter example. So as antkam says, can we apply l'Hopital's rule only if we know that $lim_{xto infty} f'(x)$ exist?
$endgroup$
– marcozz
13 hours ago






2




2




$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
13 hours ago




$begingroup$
@antkam Yes, that is correct.
$endgroup$
– José Carlos Santos
13 hours ago












$begingroup$
@marcozz Indeed.
$endgroup$
– José Carlos Santos
13 hours ago




$begingroup$
@marcozz Indeed.
$endgroup$
– José Carlos Santos
13 hours ago












$begingroup$
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
10 hours ago




$begingroup$
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
10 hours ago











3












$begingroup$

(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



$$ lim _{x to c}F(x)=lim _{xto c}G(x)=0 text{ or }pm infty, tag{1.} $$
$$ G'(x)neq 0 text{ for all }x in I, text{ with }x ne c, text{ and} tag{2.} $$
$$ lim_{x to c}frac{F'(x)}{G'(x)} text{ exists.} tag{3.} $$



then



$$lim_{x to c} frac{F(x)}{G(x)} =lim_{x to c} frac{F'(x)}{G'(x)}. tag{4.}$$



You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.



Since $lim _{xto infty}G(x)= infty$, condition $(1.)$ requires that
$$lim _{x to infty}xf(x) = infty. tag{A.}$$



Condition $(2.)$ is satisfied by $G(x)=x$.



Condition $(3.)$ requires that
$$lim_{x to infty}[f(x)+xf'(x)] text{ exists.} tag{B.}$$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    10 hours ago












  • $begingroup$
    @MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
    $endgroup$
    – steven gregory
    6 hours ago












  • $begingroup$
    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
    5 hours ago
















3












$begingroup$

(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



$$ lim _{x to c}F(x)=lim _{xto c}G(x)=0 text{ or }pm infty, tag{1.} $$
$$ G'(x)neq 0 text{ for all }x in I, text{ with }x ne c, text{ and} tag{2.} $$
$$ lim_{x to c}frac{F'(x)}{G'(x)} text{ exists.} tag{3.} $$



then



$$lim_{x to c} frac{F(x)}{G(x)} =lim_{x to c} frac{F'(x)}{G'(x)}. tag{4.}$$



You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.



Since $lim _{xto infty}G(x)= infty$, condition $(1.)$ requires that
$$lim _{x to infty}xf(x) = infty. tag{A.}$$



Condition $(2.)$ is satisfied by $G(x)=x$.



Condition $(3.)$ requires that
$$lim_{x to infty}[f(x)+xf'(x)] text{ exists.} tag{B.}$$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    10 hours ago












  • $begingroup$
    @MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
    $endgroup$
    – steven gregory
    6 hours ago












  • $begingroup$
    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
    5 hours ago














3












3








3





$begingroup$

(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



$$ lim _{x to c}F(x)=lim _{xto c}G(x)=0 text{ or }pm infty, tag{1.} $$
$$ G'(x)neq 0 text{ for all }x in I, text{ with }x ne c, text{ and} tag{2.} $$
$$ lim_{x to c}frac{F'(x)}{G'(x)} text{ exists.} tag{3.} $$



then



$$lim_{x to c} frac{F(x)}{G(x)} =lim_{x to c} frac{F'(x)}{G'(x)}. tag{4.}$$



You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.



Since $lim _{xto infty}G(x)= infty$, condition $(1.)$ requires that
$$lim _{x to infty}xf(x) = infty. tag{A.}$$



Condition $(2.)$ is satisfied by $G(x)=x$.



Condition $(3.)$ requires that
$$lim_{x to infty}[f(x)+xf'(x)] text{ exists.} tag{B.}$$



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.






share|cite|improve this answer









$endgroup$



(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



$$ lim _{x to c}F(x)=lim _{xto c}G(x)=0 text{ or }pm infty, tag{1.} $$
$$ G'(x)neq 0 text{ for all }x in I, text{ with }x ne c, text{ and} tag{2.} $$
$$ lim_{x to c}frac{F'(x)}{G'(x)} text{ exists.} tag{3.} $$



then



$$lim_{x to c} frac{F(x)}{G(x)} =lim_{x to c} frac{F'(x)}{G'(x)}. tag{4.}$$



You used $F(x) = xf(x)$ and $G(x) = x$ and $I = (x_0, infty)$ for some $x_0 < 0$.



Since $lim _{xto infty}G(x)= infty$, condition $(1.)$ requires that
$$lim _{x to infty}xf(x) = infty. tag{A.}$$



Condition $(2.)$ is satisfied by $G(x)=x$.



Condition $(3.)$ requires that
$$lim_{x to infty}[f(x)+xf'(x)] text{ exists.} tag{B.}$$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 12 hours ago









steven gregorysteven gregory

18.4k32358




18.4k32358












  • $begingroup$
    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
    10 hours ago












  • $begingroup$
    @MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
    $endgroup$
    – steven gregory
    6 hours ago












  • $begingroup$
    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
    5 hours ago


















  • $begingroup$
    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
    10 hours ago












  • $begingroup$
    @MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
    $endgroup$
    – steven gregory
    6 hours ago












  • $begingroup$
    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
    5 hours ago
















$begingroup$
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
10 hours ago






$begingroup$
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
10 hours ago














$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
6 hours ago






$begingroup$
@MarkViola Assuming he wants to use LHR on $(xf)(x)$, then since $lim_limits{x to infty} G(x) = infty$, then condition 1 requires ...
$endgroup$
– steven gregory
6 hours ago














$begingroup$
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
5 hours ago




$begingroup$
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
5 hours ago


















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