Limit of $frac{1}{7}e^{-2x^2}(1-4x^2)$ as $xtoinfty$












4














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$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$










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  • 6




    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
    yesterday












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila
    10 hours ago
















4














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$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$










share|cite|improve this question




















  • 6




    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
    yesterday












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila
    10 hours ago














4












4








4







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$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$










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






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edited 10 hours ago









Asaf Karagila

302k32426756




302k32426756










asked yesterday









B. Czostek

586




586








  • 6




    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
    yesterday












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila
    10 hours ago














  • 6




    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
    yesterday












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila
    10 hours ago








6




6




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
yesterday






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
yesterday














Is $*$ multiplication or convolution in your question?
– Asaf Karagila
10 hours ago




Is $*$ multiplication or convolution in your question?
– Asaf Karagila
10 hours ago










5 Answers
5






active

oldest

votes


















6














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



















  • 2




    L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
    – Andreas Rejbrand
    11 hours ago



















8














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































    4














    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





























      3














      Note that for $xge0$,
      $$
      begin{align}
      e^x
      &=1+x+frac{x^2}2+dots\
      &gefrac{x^2}2tag1
      end{align}
      $$

      Thus,
      $$
      e^{-2x^2}lefrac1{2x^4}tag2
      $$

      Applying $(2)$ to the expression for the derivative gives
      $$
      begin{align}
      left|frac17e^{-2x^2}!!left(1-4x^2right)right|
      &lefrac{4x^2+1}7frac1{2x^4}\
      &=frac2{7x^2}+frac1{14x^4}tag3
      end{align}
      $$






      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
          yesterday










        • @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
          – Ethan Bolker
          16 hours ago











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        5 Answers
        5






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        5 Answers
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        active

        oldest

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        active

        oldest

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        6














        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



















        • 2




          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          – Andreas Rejbrand
          11 hours ago
















        6














        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



















        • 2




          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          – Andreas Rejbrand
          11 hours ago














        6












        6








        6






        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 yesterday

























        answered yesterday









        Thomas Shelby

        1,667216




        1,667216








        • 2




          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          – Andreas Rejbrand
          11 hours ago














        • 2




          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          – Andreas Rejbrand
          11 hours ago








        2




        2




        L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
        – Andreas Rejbrand
        11 hours ago




        L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
        – Andreas Rejbrand
        11 hours ago











        8














        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




























          8














          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


























            8












            8








            8






            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 yesterday

























            answered yesterday









            KM101

            5,2691423




            5,2691423























                4














                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


























                  4














                  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
























                    4












                    4








                    4






                    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 yesterday









                    Rhys Hughes

                    4,9011427




                    4,9011427























                        3














                        Note that for $xge0$,
                        $$
                        begin{align}
                        e^x
                        &=1+x+frac{x^2}2+dots\
                        &gefrac{x^2}2tag1
                        end{align}
                        $$

                        Thus,
                        $$
                        e^{-2x^2}lefrac1{2x^4}tag2
                        $$

                        Applying $(2)$ to the expression for the derivative gives
                        $$
                        begin{align}
                        left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                        &lefrac{4x^2+1}7frac1{2x^4}\
                        &=frac2{7x^2}+frac1{14x^4}tag3
                        end{align}
                        $$






                        share|cite|improve this answer


























                          3














                          Note that for $xge0$,
                          $$
                          begin{align}
                          e^x
                          &=1+x+frac{x^2}2+dots\
                          &gefrac{x^2}2tag1
                          end{align}
                          $$

                          Thus,
                          $$
                          e^{-2x^2}lefrac1{2x^4}tag2
                          $$

                          Applying $(2)$ to the expression for the derivative gives
                          $$
                          begin{align}
                          left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                          &lefrac{4x^2+1}7frac1{2x^4}\
                          &=frac2{7x^2}+frac1{14x^4}tag3
                          end{align}
                          $$






                          share|cite|improve this answer
























                            3












                            3








                            3






                            Note that for $xge0$,
                            $$
                            begin{align}
                            e^x
                            &=1+x+frac{x^2}2+dots\
                            &gefrac{x^2}2tag1
                            end{align}
                            $$

                            Thus,
                            $$
                            e^{-2x^2}lefrac1{2x^4}tag2
                            $$

                            Applying $(2)$ to the expression for the derivative gives
                            $$
                            begin{align}
                            left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                            &lefrac{4x^2+1}7frac1{2x^4}\
                            &=frac2{7x^2}+frac1{14x^4}tag3
                            end{align}
                            $$






                            share|cite|improve this answer












                            Note that for $xge0$,
                            $$
                            begin{align}
                            e^x
                            &=1+x+frac{x^2}2+dots\
                            &gefrac{x^2}2tag1
                            end{align}
                            $$

                            Thus,
                            $$
                            e^{-2x^2}lefrac1{2x^4}tag2
                            $$

                            Applying $(2)$ to the expression for the derivative gives
                            $$
                            begin{align}
                            left|frac17e^{-2x^2}!!left(1-4x^2right)right|
                            &lefrac{4x^2+1}7frac1{2x^4}\
                            &=frac2{7x^2}+frac1{14x^4}tag3
                            end{align}
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 21 hours ago









                            robjohn

                            265k27303624




                            265k27303624























                                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
                                  yesterday










                                • @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  – Ethan Bolker
                                  16 hours 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
                                  yesterday










                                • @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  – Ethan Bolker
                                  16 hours 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 yesterday









                                Andrew

                                346115




                                346115












                                • Thank you I completely forgot the L'Hopitals Rule
                                  – B. Czostek
                                  yesterday










                                • @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  – Ethan Bolker
                                  16 hours ago


















                                • Thank you I completely forgot the L'Hopitals Rule
                                  – B. Czostek
                                  yesterday










                                • @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                  – Ethan Bolker
                                  16 hours ago
















                                Thank you I completely forgot the L'Hopitals Rule
                                – B. Czostek
                                yesterday




                                Thank you I completely forgot the L'Hopitals Rule
                                – B. Czostek
                                yesterday












                                @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                – Ethan Bolker
                                16 hours ago




                                @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
                                – Ethan Bolker
                                16 hours ago


















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