Can I conclude the following limit is infinite or it doesn't exists?












1














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



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



























    1














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



    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

























      1












      1








      1







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



      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






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      share|cite|improve this question











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      share|cite|improve this question










      asked 45 mins ago









      User

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      413311






















          2 Answers
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          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
            37 mins ago








          • 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
            36 mins ago










          • ok, will check. thanks!
            – User
            35 mins ago










          • @user376343 yes it's
            – User
            34 mins ago



















          2














          The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'H$hat{o}$pital'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























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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            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
              37 mins ago








            • 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
              36 mins ago










            • ok, will check. thanks!
              – User
              35 mins ago










            • @user376343 yes it's
              – User
              34 mins ago
















            3














            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
              37 mins ago








            • 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
              36 mins ago










            • ok, will check. thanks!
              – User
              35 mins ago










            • @user376343 yes it's
              – User
              34 mins ago














            3












            3








            3






            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 39 mins ago









            José Carlos Santos

            151k22122223




            151k22122223












            • 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
              37 mins ago








            • 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
              36 mins ago










            • ok, will check. thanks!
              – User
              35 mins ago










            • @user376343 yes it's
              – User
              34 mins ago


















            • 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
              37 mins ago








            • 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
              36 mins ago










            • ok, will check. thanks!
              – User
              35 mins ago










            • @user376343 yes it's
              – User
              34 mins ago
















            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
            37 mins ago






            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
            37 mins ago






            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
            36 mins ago




            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
            36 mins ago












            ok, will check. thanks!
            – User
            35 mins ago




            ok, will check. thanks!
            – User
            35 mins ago












            @user376343 yes it's
            – User
            34 mins ago




            @user376343 yes it's
            – User
            34 mins ago











            2














            The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'H$hat{o}$pital'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




























              2














              The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'H$hat{o}$pital'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


























                2












                2








                2






                The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'H$hat{o}$pital'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'H$hat{o}$pital'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 3 mins ago

























                answered 40 mins ago









                Kavi Rama Murthy

                50.6k31854




                50.6k31854






























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