I want to find the limit of a specific function

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$$lim_xrightarrow 0+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta$$
where $ alpha>0 $ and $ beta>0 $ are given constants






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    $$lim_xrightarrow 0+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta$$
    where $ alpha>0 $ and $ beta>0 $ are given constants






    calculus







    share|cite|improve this question









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      up vote
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      $$lim_xrightarrow 0+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta$$
      where $ alpha>0 $ and $ beta>0 $ are given constants






      calculus







      share|cite|improve this question









      New contributor




      Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $$lim_xrightarrow 0+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta$$
      where $ alpha>0 $ and $ beta>0 $ are given constants






      calculus




      calculus






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      Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited 3 hours ago









      Parcly Taxel

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          3 Answers
          3






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          Let $y=frac1x to infty$ then



          $$frace^-alphaleft(frac1x^beta-1right)x^1+beta
          =fracy^1+betae^alphaleft(y^beta-1right)
          =e^alphacdotfracy^1+betae^alpha y^betato 0$$



          indeed eventually



          $$e^alpha y^betage y^2+beta implies fracy^1+betae^alpha y^betale fracy^1+betay^2+beta=frac1y to 0$$






          share|cite|improve this answer






















          • but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
            – hamza boulahia
            3 hours ago











          • @hamzaboulahia Opsss....thanks I fix!
            – gimusi
            3 hours ago

















          up vote
          2
          down vote













          Hint. We have $alpha>0,,beta>0$ then, for some $c(alpha,beta)>0$,
          $$
          exp left(fracalphax^betaright)= sum_n=0^infty fracleft(fracalphax^betaright)^nn!ge fracc(alpha,beta)x^beta+2,qquad x>0,
          $$           giving, as $x to 0^+$,
          $$
          left|fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta right|le frace^alphax^beta+2x^beta+1c(alpha,beta)=frace^alphaxc(alpha,beta)to 0.
          $$






          share|cite|improve this answer



























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            We have,
            $$lim_xrightarrow 0^+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta=lim_xrightarrow 0+e^alphafrace^ frac-alphax^beta x^1+beta$$
            Then we substitute $dfrac1x=u$, then
            $$ lim_xrightarrow 0^+e^alphafrace^ frac-alphax^beta x^1+beta=lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta$$
            We know that the polynomial functions are negligible in front of the exponential in the vicinity of infinity,
            $$lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta=lim_urightarrow +infty e^ -alpha u^beta =0,quad mboxsince alpha>0, beta>0 $$






            share|cite|improve this answer




















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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

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              active

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              active

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              up vote
              3
              down vote













              Let $y=frac1x to infty$ then



              $$frace^-alphaleft(frac1x^beta-1right)x^1+beta
              =fracy^1+betae^alphaleft(y^beta-1right)
              =e^alphacdotfracy^1+betae^alpha y^betato 0$$



              indeed eventually



              $$e^alpha y^betage y^2+beta implies fracy^1+betae^alpha y^betale fracy^1+betay^2+beta=frac1y to 0$$






              share|cite|improve this answer






















              • but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
                – hamza boulahia
                3 hours ago











              • @hamzaboulahia Opsss....thanks I fix!
                – gimusi
                3 hours ago














              up vote
              3
              down vote













              Let $y=frac1x to infty$ then



              $$frace^-alphaleft(frac1x^beta-1right)x^1+beta
              =fracy^1+betae^alphaleft(y^beta-1right)
              =e^alphacdotfracy^1+betae^alpha y^betato 0$$



              indeed eventually



              $$e^alpha y^betage y^2+beta implies fracy^1+betae^alpha y^betale fracy^1+betay^2+beta=frac1y to 0$$






              share|cite|improve this answer






















              • but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
                – hamza boulahia
                3 hours ago











              • @hamzaboulahia Opsss....thanks I fix!
                – gimusi
                3 hours ago












              up vote
              3
              down vote










              up vote
              3
              down vote









              Let $y=frac1x to infty$ then



              $$frace^-alphaleft(frac1x^beta-1right)x^1+beta
              =fracy^1+betae^alphaleft(y^beta-1right)
              =e^alphacdotfracy^1+betae^alpha y^betato 0$$



              indeed eventually



              $$e^alpha y^betage y^2+beta implies fracy^1+betae^alpha y^betale fracy^1+betay^2+beta=frac1y to 0$$






              share|cite|improve this answer














              Let $y=frac1x to infty$ then



              $$frace^-alphaleft(frac1x^beta-1right)x^1+beta
              =fracy^1+betae^alphaleft(y^beta-1right)
              =e^alphacdotfracy^1+betae^alpha y^betato 0$$



              indeed eventually



              $$e^alpha y^betage y^2+beta implies fracy^1+betae^alpha y^betale fracy^1+betay^2+beta=frac1y to 0$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 hours ago

























              answered 3 hours ago









              gimusi

              80.7k74090




              80.7k74090











              • but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
                – hamza boulahia
                3 hours ago











              • @hamzaboulahia Opsss....thanks I fix!
                – gimusi
                3 hours ago
















              • but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
                – hamza boulahia
                3 hours ago











              • @hamzaboulahia Opsss....thanks I fix!
                – gimusi
                3 hours ago















              but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
              – hamza boulahia
              3 hours ago





              but $e^xlongrightarrow 0 $, when $xrightarrow -infty$
              – hamza boulahia
              3 hours ago













              @hamzaboulahia Opsss....thanks I fix!
              – gimusi
              3 hours ago




              @hamzaboulahia Opsss....thanks I fix!
              – gimusi
              3 hours ago










              up vote
              2
              down vote













              Hint. We have $alpha>0,,beta>0$ then, for some $c(alpha,beta)>0$,
              $$
              exp left(fracalphax^betaright)= sum_n=0^infty fracleft(fracalphax^betaright)^nn!ge fracc(alpha,beta)x^beta+2,qquad x>0,
              $$               giving, as $x to 0^+$,
              $$
              left|fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta right|le frace^alphax^beta+2x^beta+1c(alpha,beta)=frace^alphaxc(alpha,beta)to 0.
              $$






              share|cite|improve this answer
























                up vote
                2
                down vote













                Hint. We have $alpha>0,,beta>0$ then, for some $c(alpha,beta)>0$,
                $$
                exp left(fracalphax^betaright)= sum_n=0^infty fracleft(fracalphax^betaright)^nn!ge fracc(alpha,beta)x^beta+2,qquad x>0,
                $$                 giving, as $x to 0^+$,
                $$
                left|fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta right|le frace^alphax^beta+2x^beta+1c(alpha,beta)=frace^alphaxc(alpha,beta)to 0.
                $$






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Hint. We have $alpha>0,,beta>0$ then, for some $c(alpha,beta)>0$,
                  $$
                  exp left(fracalphax^betaright)= sum_n=0^infty fracleft(fracalphax^betaright)^nn!ge fracc(alpha,beta)x^beta+2,qquad x>0,
                  $$                   giving, as $x to 0^+$,
                  $$
                  left|fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta right|le frace^alphax^beta+2x^beta+1c(alpha,beta)=frace^alphaxc(alpha,beta)to 0.
                  $$






                  share|cite|improve this answer












                  Hint. We have $alpha>0,,beta>0$ then, for some $c(alpha,beta)>0$,
                  $$
                  exp left(fracalphax^betaright)= sum_n=0^infty fracleft(fracalphax^betaright)^nn!ge fracc(alpha,beta)x^beta+2,qquad x>0,
                  $$                   giving, as $x to 0^+$,
                  $$
                  left|fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta right|le frace^alphax^beta+2x^beta+1c(alpha,beta)=frace^alphaxc(alpha,beta)to 0.
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Olivier Oloa

                  107k17175293




                  107k17175293




















                      up vote
                      2
                      down vote













                      We have,
                      $$lim_xrightarrow 0^+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta=lim_xrightarrow 0+e^alphafrace^ frac-alphax^beta x^1+beta$$
                      Then we substitute $dfrac1x=u$, then
                      $$ lim_xrightarrow 0^+e^alphafrace^ frac-alphax^beta x^1+beta=lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta$$
                      We know that the polynomial functions are negligible in front of the exponential in the vicinity of infinity,
                      $$lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta=lim_urightarrow +infty e^ -alpha u^beta =0,quad mboxsince alpha>0, beta>0 $$






                      share|cite|improve this answer
























                        up vote
                        2
                        down vote













                        We have,
                        $$lim_xrightarrow 0^+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta=lim_xrightarrow 0+e^alphafrace^ frac-alphax^beta x^1+beta$$
                        Then we substitute $dfrac1x=u$, then
                        $$ lim_xrightarrow 0^+e^alphafrace^ frac-alphax^beta x^1+beta=lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta$$
                        We know that the polynomial functions are negligible in front of the exponential in the vicinity of infinity,
                        $$lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta=lim_urightarrow +infty e^ -alpha u^beta =0,quad mboxsince alpha>0, beta>0 $$






                        share|cite|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          We have,
                          $$lim_xrightarrow 0^+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta=lim_xrightarrow 0+e^alphafrace^ frac-alphax^beta x^1+beta$$
                          Then we substitute $dfrac1x=u$, then
                          $$ lim_xrightarrow 0^+e^alphafrace^ frac-alphax^beta x^1+beta=lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta$$
                          We know that the polynomial functions are negligible in front of the exponential in the vicinity of infinity,
                          $$lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta=lim_urightarrow +infty e^ -alpha u^beta =0,quad mboxsince alpha>0, beta>0 $$






                          share|cite|improve this answer












                          We have,
                          $$lim_xrightarrow 0^+fracexp left[-alphaleft( frac1x^beta -1right) right] x^1+beta=lim_xrightarrow 0+e^alphafrace^ frac-alphax^beta x^1+beta$$
                          Then we substitute $dfrac1x=u$, then
                          $$ lim_xrightarrow 0^+e^alphafrace^ frac-alphax^beta x^1+beta=lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta$$
                          We know that the polynomial functions are negligible in front of the exponential in the vicinity of infinity,
                          $$lim_urightarrow +inftye^alpha e^ -alpha u^beta u^1+beta=lim_urightarrow +infty e^ -alpha u^beta =0,quad mboxsince alpha>0, beta>0 $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          hamza boulahia

                          819316




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