How to evaluate the limit where something is raised to a power of $x$?Intuitive explanation for why...

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


Intuitive explanation for why $left(1-frac1nright)^n to frac1e$About $lim left(1+frac {x}{n}right)^n$Finding the limit of $left( 1-frac{1}{n} right)^{n}$Showing that $lim_{n rightarrow infty} left( 1 + frac{r}{n} right)^{n} = e^{r} $.Limit of $left(1+frac{a}{x}right)^x$ with and without L'Hôpital's ruleLimit of power series $ left(1+frac{x}{m}right)^m$ with m tending to infinity$Delta x$ in limit problem?Help to evaluate this limit $lim_{x to infty}x^{frac{1}{x}}$How to evaluate this exponential fraction limit?Understanding how to evaluate $lim_{xtofracpi2} frac{2^{-cos x}-1}{x-fracpi2}$How to evaluate the following complex limit?How to evaluate integrals with infinite limitsHow to evaluate $lim_{xto infty} (x^2cdotint_{0}^x e^{t^3-x^3} dt)$Evaluate limit without L'Hopital's ruleEvaluate: $lim_limits{xto3}frac{sin(x+1)}{2x(x-3)}$Using power series to evaluate a limit













7












$begingroup$


I am attempting to evaluate the following limit:



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



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 $frac{1}{e^5}$, but I am unsure how to arrive at this answer.










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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
    yesterday










  • $begingroup$
    This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
    $endgroup$
    – Xander Henderson
    17 hours ago












  • $begingroup$
    Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
    $endgroup$
    – Xander Henderson
    17 hours ago










  • $begingroup$
    It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
    $endgroup$
    – Carl Mummert
    16 hours ago
















7












$begingroup$


I am attempting to evaluate the following limit:



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



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 $frac{1}{e^5}$, but I am unsure how to arrive at this answer.










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
    yesterday










  • $begingroup$
    This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
    $endgroup$
    – Xander Henderson
    17 hours ago












  • $begingroup$
    Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
    $endgroup$
    – Xander Henderson
    17 hours ago










  • $begingroup$
    It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
    $endgroup$
    – Carl Mummert
    16 hours ago














7












7








7


0



$begingroup$


I am attempting to evaluate the following limit:



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



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 $frac{1}{e^5}$, but I am unsure how to arrive at this answer.










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:



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



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 $frac{1}{e^5}$, but I am unsure how to arrive at this answer.







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








edited 17 hours ago









Xander Henderson

14.8k103555




14.8k103555






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asked yesterday









jfkdasjfkjfkdasjfk

414




414




New contributor




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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.






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








  • 4




    $begingroup$
    Try taking the limit of the logarithm of the expression.
    $endgroup$
    – John Wayland Bales
    yesterday










  • $begingroup$
    This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
    $endgroup$
    – Xander Henderson
    17 hours ago












  • $begingroup$
    Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
    $endgroup$
    – Xander Henderson
    17 hours ago










  • $begingroup$
    It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
    $endgroup$
    – Carl Mummert
    16 hours ago














  • 4




    $begingroup$
    Try taking the limit of the logarithm of the expression.
    $endgroup$
    – John Wayland Bales
    yesterday










  • $begingroup$
    This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
    $endgroup$
    – Xander Henderson
    17 hours ago












  • $begingroup$
    Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
    $endgroup$
    – Xander Henderson
    17 hours ago










  • $begingroup$
    It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
    $endgroup$
    – Carl Mummert
    16 hours ago








4




4




$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday




$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday












$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago






$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago














$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago




$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago












$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago




$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago










6 Answers
6






active

oldest

votes


















17












$begingroup$

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
    yesterday



















9












$begingroup$

$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
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












    $begingroup$

    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









    $endgroup$





















      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 $-8$ $^{[1]}$. 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 $e$.



      [1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.






      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
        yesterday



















      1












      $begingroup$

      This is an indeterminate form $1^{infty}$ (other infinite 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
        yesterday










      • $begingroup$
        @aschepler: I meant $a^infty$. I have rephrased.
        $endgroup$
        – Yves Daoust
        21 hours ago





















      1












      $begingroup$

      A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
      $$
      lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
      $$

      If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.



      In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
      $$
      lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
      $$

      after noticing that
      $$
      frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
      $$

      This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
      $$
      h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
      $$

      the limit is $3-8=-5$.



      Alternatively, use that $log(1+u)=u+o(u)$, so you have
      $$
      lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
      lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
      $$

      Thus your limit is $e^{-5}$.






      share|cite|improve this answer









      $endgroup$













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        6 Answers
        6






        active

        oldest

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        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        17












        $begingroup$

        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
          yesterday
















        17












        $begingroup$

        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
          yesterday














        17












        17








        17





        $begingroup$

        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 yesterday









        Robert ZRobert Z

        99.6k1068140




        99.6k1068140












        • $begingroup$
          Thank you this helped a lot!
          $endgroup$
          – jfkdasjfk
          yesterday


















        • $begingroup$
          Thank you this helped a lot!
          $endgroup$
          – jfkdasjfk
          yesterday
















        $begingroup$
        Thank you this helped a lot!
        $endgroup$
        – jfkdasjfk
        yesterday




        $begingroup$
        Thank you this helped a lot!
        $endgroup$
        – jfkdasjfk
        yesterday











        9












        $begingroup$

        $$=lim_{xto infty} (1-frac{5}{x})^x$$
        $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
        $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
        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









        $endgroup$


















          9












          $begingroup$

          $$=lim_{xto infty} (1-frac{5}{x})^x$$
          $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
          $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
          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









          $endgroup$
















            9












            9








            9





            $begingroup$

            $$=lim_{xto infty} (1-frac{5}{x})^x$$
            $$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
            $$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
            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 yesterday









            Peter ForemanPeter Foreman

            2,9071214




            2,9071214























                2












                $begingroup$

                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









                $endgroup$


















                  2












                  $begingroup$

                  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









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    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 yesterday









                    Paras KhoslaParas Khosla

                    1,648219




                    1,648219























                        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 $-8$ $^{[1]}$. 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 $e$.



                        [1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.






                        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
                          yesterday
















                        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 $-8$ $^{[1]}$. 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 $e$.



                        [1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.






                        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
                          yesterday














                        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 $-8$ $^{[1]}$. 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 $e$.



                        [1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.






                        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 $-8$ $^{[1]}$. 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 $e$.



                        [1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 14 hours ago

























                        answered yesterday









                        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
                          yesterday


















                        • $begingroup$
                          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
                          yesterday
















                        $begingroup$
                        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
                        yesterday




                        $begingroup$
                        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
                        yesterday











                        1












                        $begingroup$

                        This is an indeterminate form $1^{infty}$ (other infinite 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
                          yesterday










                        • $begingroup$
                          @aschepler: I meant $a^infty$. I have rephrased.
                          $endgroup$
                          – Yves Daoust
                          21 hours ago


















                        1












                        $begingroup$

                        This is an indeterminate form $1^{infty}$ (other infinite 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
                          yesterday










                        • $begingroup$
                          @aschepler: I meant $a^infty$. I have rephrased.
                          $endgroup$
                          – Yves Daoust
                          21 hours ago
















                        1












                        1








                        1





                        $begingroup$

                        This is an indeterminate form $1^{infty}$ (other infinite 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 infinite 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








                        edited 21 hours ago

























                        answered yesterday









                        Yves DaoustYves Daoust

                        129k676227




                        129k676227












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










                        • $begingroup$
                          @aschepler: I meant $a^infty$. I have rephrased.
                          $endgroup$
                          – Yves Daoust
                          21 hours ago




















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










                        • $begingroup$
                          @aschepler: I meant $a^infty$. I have rephrased.
                          $endgroup$
                          – Yves Daoust
                          21 hours ago


















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




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












                        $begingroup$
                        @aschepler: I meant $a^infty$. I have rephrased.
                        $endgroup$
                        – Yves Daoust
                        21 hours ago






                        $begingroup$
                        @aschepler: I meant $a^infty$. I have rephrased.
                        $endgroup$
                        – Yves Daoust
                        21 hours ago













                        1












                        $begingroup$

                        A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
                        $$
                        lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
                        $$

                        If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.



                        In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
                        $$
                        lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
                        $$

                        after noticing that
                        $$
                        frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
                        $$

                        This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
                        $$
                        h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
                        $$

                        the limit is $3-8=-5$.



                        Alternatively, use that $log(1+u)=u+o(u)$, so you have
                        $$
                        lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
                        lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
                        $$

                        Thus your limit is $e^{-5}$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
                          $$
                          lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
                          $$

                          If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.



                          In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
                          $$
                          lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
                          $$

                          after noticing that
                          $$
                          frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
                          $$

                          This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
                          $$
                          h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
                          $$

                          the limit is $3-8=-5$.



                          Alternatively, use that $log(1+u)=u+o(u)$, so you have
                          $$
                          lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
                          lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
                          $$

                          Thus your limit is $e^{-5}$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
                            $$
                            lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
                            $$

                            If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.



                            In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
                            $$
                            lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
                            $$

                            after noticing that
                            $$
                            frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
                            $$

                            This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
                            $$
                            h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
                            $$

                            the limit is $3-8=-5$.



                            Alternatively, use that $log(1+u)=u+o(u)$, so you have
                            $$
                            lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
                            lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
                            $$

                            Thus your limit is $e^{-5}$.






                            share|cite|improve this answer









                            $endgroup$



                            A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
                            $$
                            lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
                            $$

                            If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.



                            In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
                            $$
                            lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
                            $$

                            after noticing that
                            $$
                            frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
                            $$

                            This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
                            $$
                            h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
                            $$

                            the limit is $3-8=-5$.



                            Alternatively, use that $log(1+u)=u+o(u)$, so you have
                            $$
                            lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
                            lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
                            $$

                            Thus your limit is $e^{-5}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 19 hours ago









                            egregegreg

                            183k1486205




                            183k1486205






















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