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
    Jan 3 at 15:45












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila♦
    Jan 4 at 7:50
















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
    Jan 3 at 15:45












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila♦
    Jan 4 at 7:50














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 Jan 4 at 7:51









Asaf Karagila♦

302k32427757




302k32427757










asked Jan 3 at 15:25









B. CzostekB. Czostek

566




566








  • 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
    Jan 3 at 15:45












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila♦
    Jan 4 at 7:50














  • 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
    Jan 3 at 15:45












  • Is $*$ multiplication or convolution in your question?
    – Asaf Karagila♦
    Jan 4 at 7:50








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
Jan 3 at 15:45






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
Jan 3 at 15:45














Is $*$ multiplication or convolution in your question?
– Asaf Karagila♦
Jan 4 at 7:50




Is $*$ multiplication or convolution in your question?
– Asaf Karagila♦
Jan 4 at 7:50










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
    Jan 4 at 7:01



















9














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































    5














    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














      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
          Jan 3 at 16:04










        • @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
          Jan 4 at 1:34











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        5 Answers
        5






        active

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        5 Answers
        5






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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
          Jan 4 at 7:01
















        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
          Jan 4 at 7:01














        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 Jan 3 at 15:39

























        answered Jan 3 at 15:32









        Thomas ShelbyThomas Shelby

        1,893217




        1,893217








        • 2




          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          – Andreas Rejbrand
          Jan 4 at 7:01














        • 2




          L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
          – Andreas Rejbrand
          Jan 4 at 7:01








        2




        2




        L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
        – Andreas Rejbrand
        Jan 4 at 7:01




        L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
        – Andreas Rejbrand
        Jan 4 at 7:01











        9














        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




























          9














          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


























            9












            9








            9






            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 Jan 3 at 17:17

























            answered Jan 3 at 16:00









            KM101KM101

            5,5611423




            5,5611423























                5














                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


























                  5














                  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
























                    5












                    5








                    5






                    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 Jan 3 at 16:49









                    Rhys HughesRhys Hughes

                    5,0441427




                    5,0441427























                        4














                        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


























                          4














                          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
























                            4












                            4








                            4






                            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 Jan 3 at 20:10









                            robjohn♦robjohn

                            265k27303625




                            265k27303625























                                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
                                  Jan 3 at 16:04










                                • @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
                                  Jan 4 at 1:34
















                                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
                                  Jan 3 at 16:04










                                • @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
                                  Jan 4 at 1:34














                                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 Jan 3 at 15:42









                                AndrewAndrew

                                346115




                                346115












                                • Thank you I completely forgot the L'Hopitals Rule
                                  – B. Czostek
                                  Jan 3 at 16:04










                                • @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
                                  Jan 4 at 1:34


















                                • Thank you I completely forgot the L'Hopitals Rule
                                  – B. Czostek
                                  Jan 3 at 16:04










                                • @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
                                  Jan 4 at 1:34
















                                Thank you I completely forgot the L'Hopitals Rule
                                – B. Czostek
                                Jan 3 at 16:04




                                Thank you I completely forgot the L'Hopitals Rule
                                – B. Czostek
                                Jan 3 at 16:04












                                @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
                                Jan 4 at 1:34




                                @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
                                Jan 4 at 1:34


















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                                1300-talet

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                                Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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                                Display a custom attribute below product name in the front-end Magento 1.9.3.8

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                                0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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