How to evaluate the limit where something is raised to a power of x?












7












$begingroup$


I am attempting to evaluate the following limit:



$$lim_{xto infty} Biggl({x+3over x+8}Biggl)^x$$



I was wondering if anyone could share some strategies for evaluating limits raised to a power of x, as I have never encountered these before.



I have found the answer to be ${1over e^5}$, but I am unsure how to arrive at this answer.



Thank you.






calculus limits







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  • 4




    $begingroup$
    Try taking the limit of the logarithm of the expression.
    $endgroup$
    – John Wayland Bales
    13 hours ago
















7












$begingroup$


I am attempting to evaluate the following limit:



$$lim_{xto infty} Biggl({x+3over x+8}Biggl)^x$$



I was wondering if anyone could share some strategies for evaluating limits raised to a power of x, as I have never encountered these before.



I have found the answer to be ${1over e^5}$, but I am unsure how to arrive at this answer.



Thank you.






calculus limits







share|cite|improve this question







New contributor




jfkdasjfk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    Try taking the limit of the logarithm of the expression.
    $endgroup$
    – John Wayland Bales
    13 hours ago














7












7








7


0



$begingroup$


I am attempting to evaluate the following limit:



$$lim_{xto infty} Biggl({x+3over x+8}Biggl)^x$$



I was wondering if anyone could share some strategies for evaluating limits raised to a power of x, as I have never encountered these before.



I have found the answer to be ${1over e^5}$, but I am unsure how to arrive at this answer.



Thank you.






calculus limits







share|cite|improve this question







New contributor




jfkdasjfk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am attempting to evaluate the following limit:



$$lim_{xto infty} Biggl({x+3over x+8}Biggl)^x$$



I was wondering if anyone could share some strategies for evaluating limits raised to a power of x, as I have never encountered these before.



I have found the answer to be ${1over e^5}$, but I am unsure how to arrive at this answer.



Thank you.






calculus limits




calculus limits






share|cite|improve this question







New contributor




jfkdasjfk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




jfkdasjfk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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asked 13 hours ago









jfkdasjfkjfkdasjfk

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383




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New contributor





jfkdasjfk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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jfkdasjfk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 4




    $begingroup$
    Try taking the limit of the logarithm of the expression.
    $endgroup$
    – John Wayland Bales
    13 hours ago














  • 4




    $begingroup$
    Try taking the limit of the logarithm of the expression.
    $endgroup$
    – John Wayland Bales
    13 hours ago








4




4




$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
13 hours ago




$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
13 hours ago










5 Answers
5






active

oldest

votes


















14












$begingroup$

Hint. Note that
$$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you this helped a lot!
    $endgroup$
    – jfkdasjfk
    12 hours ago



















8












$begingroup$

$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
Now in order to evaluate
$$=lim_{xto infty} xln{(1-frac{5}{x})}$$
$$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
One can use L'Hôpitals rule giving
$$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
$$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
$$=-5$$
Hence the initial limit is
$$e^{-5}=frac{1}{e^5}$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
      $endgroup$
      – Pedro
      4 hours ago



















    1












    $begingroup$

    Hint:



    Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.



    $$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      This is an indeterminate form $1^{infty}$ (other powers usually Raise no difficulty).



      Writing the limit as



      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have



      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
      \=e^{lim_{xtoinfty}x/f(x)}.$$




      In the given case, we have $f(x)=-(x+8)/5.$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        What about "other powers" with the form "$infty^0$"?
        $endgroup$
        – aschepler
        5 hours ago











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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      14












      $begingroup$

      Hint. Note that
      $$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
      Moreover, for $anot=0$, after letting $t=x/a$ we have that
      $$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
      where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you this helped a lot!
        $endgroup$
        – jfkdasjfk
        12 hours ago
















      14












      $begingroup$

      Hint. Note that
      $$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
      Moreover, for $anot=0$, after letting $t=x/a$ we have that
      $$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
      where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you this helped a lot!
        $endgroup$
        – jfkdasjfk
        12 hours ago














      14












      14








      14





      $begingroup$

      Hint. Note that
      $$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
      Moreover, for $anot=0$, after letting $t=x/a$ we have that
      $$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
      where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.






      share|cite|improve this answer









      $endgroup$



      Hint. Note that
      $$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
      Moreover, for $anot=0$, after letting $t=x/a$ we have that
      $$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
      where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 12 hours ago









      Robert ZRobert Z

      99.4k1068140




      99.4k1068140












      • $begingroup$
        Thank you this helped a lot!
        $endgroup$
        – jfkdasjfk
        12 hours ago


















      • $begingroup$
        Thank you this helped a lot!
        $endgroup$
        – jfkdasjfk
        12 hours ago
















      $begingroup$
      Thank you this helped a lot!
      $endgroup$
      – jfkdasjfk
      12 hours ago




      $begingroup$
      Thank you this helped a lot!
      $endgroup$
      – jfkdasjfk
      12 hours ago











      8












      $begingroup$

      $$=lim_{xto infty} (1-frac{5}{x})^x$$
      $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
      $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
      Now in order to evaluate
      $$=lim_{xto infty} xln{(1-frac{5}{x})}$$
      $$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
      One can use L'Hôpitals rule giving
      $$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
      $$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
      $$=-5$$
      Hence the initial limit is
      $$e^{-5}=frac{1}{e^5}$$






      share|cite|improve this answer









      $endgroup$


















        8












        $begingroup$

        $$=lim_{xto infty} (1-frac{5}{x})^x$$
        $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
        $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
        Now in order to evaluate
        $$=lim_{xto infty} xln{(1-frac{5}{x})}$$
        $$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
        One can use L'Hôpitals rule giving
        $$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
        $$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
        $$=-5$$
        Hence the initial limit is
        $$e^{-5}=frac{1}{e^5}$$






        share|cite|improve this answer









        $endgroup$
















          8












          8








          8





          $begingroup$

          $$=lim_{xto infty} (1-frac{5}{x})^x$$
          $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
          $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
          Now in order to evaluate
          $$=lim_{xto infty} xln{(1-frac{5}{x})}$$
          $$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
          One can use L'Hôpitals rule giving
          $$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
          $$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
          $$=-5$$
          Hence the initial limit is
          $$e^{-5}=frac{1}{e^5}$$






          share|cite|improve this answer









          $endgroup$



          $$=lim_{xto infty} (1-frac{5}{x})^x$$
          $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
          $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
          Now in order to evaluate
          $$=lim_{xto infty} xln{(1-frac{5}{x})}$$
          $$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
          One can use L'Hôpitals rule giving
          $$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
          $$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
          $$=-5$$
          Hence the initial limit is
          $$e^{-5}=frac{1}{e^5}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 12 hours ago









          Peter ForemanPeter Foreman

          2,8871214




          2,8871214























              2












              $begingroup$

              Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
                $endgroup$
                – Pedro
                4 hours ago
















              2












              $begingroup$

              Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
                $endgroup$
                – Pedro
                4 hours ago














              2












              2








              2





              $begingroup$

              Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.






              share|cite|improve this answer









              $endgroup$



              Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 7 hours ago









              AcccumulationAcccumulation

              7,0852619




              7,0852619












              • $begingroup$
                In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
                $endgroup$
                – Pedro
                4 hours ago


















              • $begingroup$
                In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
                $endgroup$
                – Pedro
                4 hours ago
















              $begingroup$
              In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
              $endgroup$
              – Pedro
              4 hours ago




              $begingroup$
              In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
              $endgroup$
              – Pedro
              4 hours ago











              1












              $begingroup$

              Hint:



              Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.



              $$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint:



                Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.



                $$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint:



                  Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.



                  $$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.



                  $$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 12 hours ago









                  Paras KhoslaParas Khosla

                  1,627219




                  1,627219























                      1












                      $begingroup$

                      This is an indeterminate form $1^{infty}$ (other powers usually Raise no difficulty).



                      Writing the limit as



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
                      \=e^{lim_{xtoinfty}x/f(x)}.$$




                      In the given case, we have $f(x)=-(x+8)/5.$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        What about "other powers" with the form "$infty^0$"?
                        $endgroup$
                        – aschepler
                        5 hours ago
















                      1












                      $begingroup$

                      This is an indeterminate form $1^{infty}$ (other powers usually Raise no difficulty).



                      Writing the limit as



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
                      \=e^{lim_{xtoinfty}x/f(x)}.$$




                      In the given case, we have $f(x)=-(x+8)/5.$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        What about "other powers" with the form "$infty^0$"?
                        $endgroup$
                        – aschepler
                        5 hours ago














                      1












                      1








                      1





                      $begingroup$

                      This is an indeterminate form $1^{infty}$ (other powers usually Raise no difficulty).



                      Writing the limit as



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
                      \=e^{lim_{xtoinfty}x/f(x)}.$$




                      In the given case, we have $f(x)=-(x+8)/5.$






                      share|cite|improve this answer









                      $endgroup$



                      This is an indeterminate form $1^{infty}$ (other powers usually Raise no difficulty).



                      Writing the limit as



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have



                      $$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
                      \=e^{lim_{xtoinfty}x/f(x)}.$$




                      In the given case, we have $f(x)=-(x+8)/5.$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 7 hours ago









                      Yves DaoustYves Daoust

                      129k676227




                      129k676227












                      • $begingroup$
                        What about "other powers" with the form "$infty^0$"?
                        $endgroup$
                        – aschepler
                        5 hours ago


















                      • $begingroup$
                        What about "other powers" with the form "$infty^0$"?
                        $endgroup$
                        – aschepler
                        5 hours ago
















                      $begingroup$
                      What about "other powers" with the form "$infty^0$"?
                      $endgroup$
                      – aschepler
                      5 hours ago




                      $begingroup$
                      What about "other powers" with the form "$infty^0$"?
                      $endgroup$
                      – aschepler
                      5 hours ago










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