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






limits exponential-function







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






limits exponential-function







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






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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share|cite|improve this question













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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






        active

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        5 Answers
        5






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        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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                                Has there ever been an instance of an active nuclear power plant within or near a war zone?

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                                36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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