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














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



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




















  • 2




    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?
    – User
    yesterday
















3














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



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




















  • 2




    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?
    – User
    yesterday














3












3








3







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



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















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



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 yesterday









Asaf Karagila♦

302k32426756




302k32426756










asked yesterday









User

425311




425311








  • 2




    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?
    – User
    yesterday














  • 2




    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?
    – User
    yesterday








2




2




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?
– User
yesterday




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?
– User
yesterday










2 Answers
2






active

oldest

votes


















6














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





















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday



















4














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























  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














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





















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday
















6














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





















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday














6












6








6






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












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 yesterday









José Carlos Santos

151k22123224




151k22123224












  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday


















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday
















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?
– User
yesterday






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?
– User
yesterday






1




1




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.
– José Carlos Santos
yesterday




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.
– José Carlos Santos
yesterday












ok, will check. thanks!
– User
yesterday




ok, will check. thanks!
– User
yesterday












@user376343 yes it's
– User
yesterday




@user376343 yes it's
– User
yesterday











4














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























  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday


















4














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























  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday
















4












4








4






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














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 yesterday









Surb

37.4k94375




37.4k94375










answered yesterday









Kavi Rama Murthy

51k31854




51k31854












  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday




















  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday


















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 $infty$...
– User
yesterday




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 $infty$...
– User
yesterday












@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.
– Kavi Rama Murthy
yesterday






@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.
– Kavi Rama Murthy
yesterday




















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