Can I conclude that $lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$ is infinite or it doesn't...












3












$begingroup$


$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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







share|cite|improve this question











$endgroup$








  • 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
















3












$begingroup$


$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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







share|cite|improve this question











$endgroup$








  • 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














3












3








3





$begingroup$


$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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







share|cite|improve this question











$endgroup$




$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 13:21









Asaf Karagila♦

307k33440773




307k33440773










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














  • 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










2 Answers
2






active

oldest

votes


















6












$begingroup$

The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















4












$begingroup$

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 $|cos (t)| leq 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















6












$begingroup$

The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














6












6








6





$begingroup$

The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.






share|cite|improve this answer









$endgroup$



The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 9:51









José Carlos SantosJosé Carlos Santos

172k23132240




172k23132240












  • $begingroup$
    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








  • 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











4












$begingroup$

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 $|cos (t)| leq 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















4












$begingroup$

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 $|cos (t)| leq 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















4












4








4





$begingroup$

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 $|cos (t)| leq 1$.






share|cite|improve this answer











$endgroup$



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 $|cos (t)| leq 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 11:46









Surb

38.4k94478




38.4k94478










answered Jan 3 at 9:50









Kavi Rama MurthyKavi Rama Murthy

71.8k53170




71.8k53170












  • $begingroup$
    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










  • $begingroup$
    @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




















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