Can I conclude that $lim_xto0^+fracx^2e^-frac1x^2cos(frac1x^2)^2$ is infinite or it doesn't exists?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
$$lim_xto0^+fracx^2e^-frac1x^2cos(frac1x^2)^2$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
asked Jan 3 at 9:45
UserUser
452312
$endgroup$
add a comment |
$begingroup$
$$lim_xto0^+fracx^2e^-frac1x^2cos(frac1x^2)^2$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
asked Jan 3 at 9:45
UserUser
452312
$endgroup$
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
add a comment |
$begingroup$
$$lim_xto0^+fracx^2e^-frac1x^2cos(frac1x^2)^2$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
asked Jan 3 at 9:45
UserUser
452312
$endgroup$
$$lim_xto0^+fracx^2e^-frac1x^2cos(frac1x^2)^2$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
asked Jan 3 at 9:45
UserUser
452312
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
asked Jan 3 at 9:45
UserUser
452312
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
edited Jan 3 at 13:21
Asaf Karagila♦
302k32429760
302k32429760
asked Jan 3 at 9:45
UserUser
452312
asked Jan 3 at 9:45
UserUser
452312
asked Jan 3 at 9:45
UserUser
452312
452312
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
add a comment |
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
2
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The limit$$lim_xto0^+fracx^2expleft(-frac1x^2right)$$is indeed $+infty$. Since $dfrac1cos^2left(frac1x^2right)geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac e^y y^2 to infty$ as $y to infty$. Replace $y$ by $frac 1 x^2$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
37.5k94375
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
The limit$$lim_xto0^+fracx^2expleft(-frac1x^2right)$$is indeed $+infty$. Since $dfrac1cos^2left(frac1x^2right)geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The limit$$lim_xto0^+fracx^2expleft(-frac1x^2right)$$is indeed $+infty$. Since $dfrac1cos^2left(frac1x^2right)geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The limit$$lim_xto0^+fracx^2expleft(-frac1x^2right)$$is indeed $+infty$. Since $dfrac1cos^2left(frac1x^2right)geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
The limit$$lim_xto0^+fracx^2expleft(-frac1x^2right)$$is indeed $+infty$. Since $dfrac1cos^2left(frac1x^2right)geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
154k22124227
154k22124227
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac e^y y^2 to infty$ as $y to infty$. Replace $y$ by $frac 1 x^2$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
37.5k94375
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac e^y y^2 to infty$ as $y to infty$. Replace $y$ by $frac 1 x^2$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
37.5k94375
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac e^y y^2 to infty$ as $y to infty$. Replace $y$ by $frac 1 x^2$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
37.5k94375
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac e^y y^2 to infty$ as $y to infty$. Replace $y$ by $frac 1 x^2$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
37.5k94375
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
edited Jan 3 at 11:46
Surb
37.5k94375
edited Jan 3 at 11:46
Surb
37.5k94375
edited Jan 3 at 11:46
Surb
37.5k94375
37.5k94375
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
53.8k32055
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
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$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02