Limit of function with Euler's Number in unfamiliar form












1














I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function







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  • 2




    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    – Doug M
    1 hour ago


















1














I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function







share|cite|improve this question


















  • 2




    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    – Doug M
    1 hour ago
















1












1








1







I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function







share|cite|improve this question













I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)



I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$






limits exponential-function




limits exponential-function






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asked 2 hours ago









B. Czostek

366




366








  • 2




    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    – Doug M
    1 hour ago
















  • 2




    Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
    – Doug M
    1 hour ago










2




2




Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
– Doug M
1 hour ago






Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
– Doug M
1 hour ago












4 Answers
4






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3














Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.






share|cite|improve this answer































    3














    Hint: Rewrite the expression as



    $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



    Now, notice the growth of the numerator and denominator. Which grows more quickly?






    share|cite|improve this answer































      2














      You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



      I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
      $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






      share|cite|improve this answer





















      • Thank you I completely forgot the L'Hopitals Rule
        – B. Czostek
        1 hour ago



















      1














      Your derivative is correct.



      Then arrange it as:



      $$frac{1-4x^2}{7e^{2x^2}}$$



      Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3














        Using L'Hôpital's rule , we get
        begin{align}
        lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
        &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
        end{align}        .






        share|cite|improve this answer




























          3














          Using L'Hôpital's rule , we get
          begin{align}
          lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
          &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
          end{align}          .






          share|cite|improve this answer


























            3












            3








            3






            Using L'Hôpital's rule , we get
            begin{align}
            lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
            &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
            end{align}            .






            share|cite|improve this answer














            Using L'Hôpital's rule , we get
            begin{align}
            lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
            &=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
            end{align}            .







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            Thomas Shelby

            1,617216




            1,617216























                3














                Hint: Rewrite the expression as



                $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



                Now, notice the growth of the numerator and denominator. Which grows more quickly?






                share|cite|improve this answer




























                  3














                  Hint: Rewrite the expression as



                  $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



                  Now, notice the growth of the numerator and denominator. Which grows more quickly?






                  share|cite|improve this answer


























                    3












                    3








                    3






                    Hint: Rewrite the expression as



                    $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



                    Now, notice the growth of the numerator and denominator. Which grows more quickly?






                    share|cite|improve this answer














                    Hint: Rewrite the expression as



                    $$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$



                    Now, notice the growth of the numerator and denominator. Which grows more quickly?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 24 mins ago

























                    answered 1 hour ago









                    KM101

                    5,0791423




                    5,0791423























                        2














                        You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                        I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                        $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






                        share|cite|improve this answer





















                        • Thank you I completely forgot the L'Hopitals Rule
                          – B. Czostek
                          1 hour ago
















                        2














                        You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                        I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                        $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






                        share|cite|improve this answer





















                        • Thank you I completely forgot the L'Hopitals Rule
                          – B. Czostek
                          1 hour ago














                        2












                        2








                        2






                        You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                        I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                        $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.






                        share|cite|improve this answer












                        You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$



                        I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
                        $$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 1 hour ago









                        Andrew

                        346115




                        346115












                        • Thank you I completely forgot the L'Hopitals Rule
                          – B. Czostek
                          1 hour ago


















                        • Thank you I completely forgot the L'Hopitals Rule
                          – B. Czostek
                          1 hour ago
















                        Thank you I completely forgot the L'Hopitals Rule
                        – B. Czostek
                        1 hour ago




                        Thank you I completely forgot the L'Hopitals Rule
                        – B. Czostek
                        1 hour ago











                        1














                        Your derivative is correct.



                        Then arrange it as:



                        $$frac{1-4x^2}{7e^{2x^2}}$$



                        Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






                        share|cite|improve this answer


























                          1














                          Your derivative is correct.



                          Then arrange it as:



                          $$frac{1-4x^2}{7e^{2x^2}}$$



                          Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Your derivative is correct.



                            Then arrange it as:



                            $$frac{1-4x^2}{7e^{2x^2}}$$



                            Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$






                            share|cite|improve this answer












                            Your derivative is correct.



                            Then arrange it as:



                            $$frac{1-4x^2}{7e^{2x^2}}$$



                            Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 52 mins ago









                            Rhys Hughes

                            4,8261427




                            4,8261427






























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