Limit $lim_{xtoinfty}left(1-frac{a^2}{x^2}right)^{x^2}$












1












$begingroup$


I found this example in a textbook, and I understand the author's reasoning and I also reached the same answer using L’Hôpital’s rule. However, I have two issues:



Firstly: For any finite $a$, then as $x rightarrow infty$, $1-frac{a^2}{x^2} rightarrow1$, and 1 to the power of anything is 1, so shouldn't the answer be 1?



Secondly: If I try some form of estimation on my calculator by replacing $x^2$ with $10^{99}$, I also get 1 for any $a$ I use.










share|cite|improve this question











$endgroup$












  • $begingroup$
    See brilliant.org/wiki/indeterminate-forms
    $endgroup$
    – lab bhattacharjee
    Dec 9 '18 at 10:12










  • $begingroup$
    This really looks like a variation on $e^{-a^2}$
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:48












  • $begingroup$
    Also see wolframalpha.com/input/…
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:58






  • 1




    $begingroup$
    See also: About $lim left(1+frac {x}{n}right)^n$ and other posts linked there. (Notice that after substitution $t=x^2$, you limit is $lim_{ttoinfty} left(1-frac{a^2}tright)^t$.
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 11:04












  • $begingroup$
    I think $10^{99}$ might be a bit of an overkill for a computer to not start rounding off. I would try smaller yet still large values like $10^3$, $10^5$, $10^{10}$, this will still give you the convergence you desire, but without computational rounding errors.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 11:27


















1












$begingroup$


I found this example in a textbook, and I understand the author's reasoning and I also reached the same answer using L’Hôpital’s rule. However, I have two issues:



Firstly: For any finite $a$, then as $x rightarrow infty$, $1-frac{a^2}{x^2} rightarrow1$, and 1 to the power of anything is 1, so shouldn't the answer be 1?



Secondly: If I try some form of estimation on my calculator by replacing $x^2$ with $10^{99}$, I also get 1 for any $a$ I use.










share|cite|improve this question











$endgroup$












  • $begingroup$
    See brilliant.org/wiki/indeterminate-forms
    $endgroup$
    – lab bhattacharjee
    Dec 9 '18 at 10:12










  • $begingroup$
    This really looks like a variation on $e^{-a^2}$
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:48












  • $begingroup$
    Also see wolframalpha.com/input/…
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:58






  • 1




    $begingroup$
    See also: About $lim left(1+frac {x}{n}right)^n$ and other posts linked there. (Notice that after substitution $t=x^2$, you limit is $lim_{ttoinfty} left(1-frac{a^2}tright)^t$.
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 11:04












  • $begingroup$
    I think $10^{99}$ might be a bit of an overkill for a computer to not start rounding off. I would try smaller yet still large values like $10^3$, $10^5$, $10^{10}$, this will still give you the convergence you desire, but without computational rounding errors.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 11:27
















1












1








1


1



$begingroup$


I found this example in a textbook, and I understand the author's reasoning and I also reached the same answer using L’Hôpital’s rule. However, I have two issues:



Firstly: For any finite $a$, then as $x rightarrow infty$, $1-frac{a^2}{x^2} rightarrow1$, and 1 to the power of anything is 1, so shouldn't the answer be 1?



Secondly: If I try some form of estimation on my calculator by replacing $x^2$ with $10^{99}$, I also get 1 for any $a$ I use.










share|cite|improve this question











$endgroup$




I found this example in a textbook, and I understand the author's reasoning and I also reached the same answer using L’Hôpital’s rule. However, I have two issues:



Firstly: For any finite $a$, then as $x rightarrow infty$, $1-frac{a^2}{x^2} rightarrow1$, and 1 to the power of anything is 1, so shouldn't the answer be 1?



Secondly: If I try some form of estimation on my calculator by replacing $x^2$ with $10^{99}$, I also get 1 for any $a$ I use.







limits taylor-expansion infinity estimation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 10:59









Martin Sleziak

44.7k9117272




44.7k9117272










asked Dec 9 '18 at 10:10









o co c

171




171












  • $begingroup$
    See brilliant.org/wiki/indeterminate-forms
    $endgroup$
    – lab bhattacharjee
    Dec 9 '18 at 10:12










  • $begingroup$
    This really looks like a variation on $e^{-a^2}$
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:48












  • $begingroup$
    Also see wolframalpha.com/input/…
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:58






  • 1




    $begingroup$
    See also: About $lim left(1+frac {x}{n}right)^n$ and other posts linked there. (Notice that after substitution $t=x^2$, you limit is $lim_{ttoinfty} left(1-frac{a^2}tright)^t$.
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 11:04












  • $begingroup$
    I think $10^{99}$ might be a bit of an overkill for a computer to not start rounding off. I would try smaller yet still large values like $10^3$, $10^5$, $10^{10}$, this will still give you the convergence you desire, but without computational rounding errors.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 11:27




















  • $begingroup$
    See brilliant.org/wiki/indeterminate-forms
    $endgroup$
    – lab bhattacharjee
    Dec 9 '18 at 10:12










  • $begingroup$
    This really looks like a variation on $e^{-a^2}$
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:48












  • $begingroup$
    Also see wolframalpha.com/input/…
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 10:58






  • 1




    $begingroup$
    See also: About $lim left(1+frac {x}{n}right)^n$ and other posts linked there. (Notice that after substitution $t=x^2$, you limit is $lim_{ttoinfty} left(1-frac{a^2}tright)^t$.
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 11:04












  • $begingroup$
    I think $10^{99}$ might be a bit of an overkill for a computer to not start rounding off. I would try smaller yet still large values like $10^3$, $10^5$, $10^{10}$, this will still give you the convergence you desire, but without computational rounding errors.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 11:27


















$begingroup$
See brilliant.org/wiki/indeterminate-forms
$endgroup$
– lab bhattacharjee
Dec 9 '18 at 10:12




$begingroup$
See brilliant.org/wiki/indeterminate-forms
$endgroup$
– lab bhattacharjee
Dec 9 '18 at 10:12












$begingroup$
This really looks like a variation on $e^{-a^2}$
$endgroup$
– Wesley Strik
Dec 9 '18 at 10:48






$begingroup$
This really looks like a variation on $e^{-a^2}$
$endgroup$
– Wesley Strik
Dec 9 '18 at 10:48














$begingroup$
Also see wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Dec 9 '18 at 10:58




$begingroup$
Also see wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Dec 9 '18 at 10:58




1




1




$begingroup$
See also: About $lim left(1+frac {x}{n}right)^n$ and other posts linked there. (Notice that after substitution $t=x^2$, you limit is $lim_{ttoinfty} left(1-frac{a^2}tright)^t$.
$endgroup$
– Martin Sleziak
Dec 9 '18 at 11:04






$begingroup$
See also: About $lim left(1+frac {x}{n}right)^n$ and other posts linked there. (Notice that after substitution $t=x^2$, you limit is $lim_{ttoinfty} left(1-frac{a^2}tright)^t$.
$endgroup$
– Martin Sleziak
Dec 9 '18 at 11:04














$begingroup$
I think $10^{99}$ might be a bit of an overkill for a computer to not start rounding off. I would try smaller yet still large values like $10^3$, $10^5$, $10^{10}$, this will still give you the convergence you desire, but without computational rounding errors.
$endgroup$
– Wesley Strik
Dec 9 '18 at 11:27






$begingroup$
I think $10^{99}$ might be a bit of an overkill for a computer to not start rounding off. I would try smaller yet still large values like $10^3$, $10^5$, $10^{10}$, this will still give you the convergence you desire, but without computational rounding errors.
$endgroup$
– Wesley Strik
Dec 9 '18 at 11:27












4 Answers
4






active

oldest

votes


















2












$begingroup$

We are using that



$$left(1-frac{a^2}{x^2}right)^{x^2}=e^{x^2 lnleft(1-frac{a^2}{x^2}right)}$$



and $x^2 lnleft(1-frac{a^2}{x^2}right)$ is an indeterminate form $infty cdot 0$ and since the Taylor's expansion for $ln (1+t)$ as $tto 0$ is



$$ln (1+t)=t-frac12t^2+frac13 t^3-frac14t^4+ldots$$



the result follows.



As an intuitive explanation of the result and for the indeterminate form $1^{infty}$, we need to consider that, in that case, while the base tends to $1^-$ the effect of the exponent is to make the quantity smaller towards zero. The final result is indeed a value in between that is $frac1{e^{a^2}}$.



That's of course not always the case since all depends upon the rate of convergence to $1$ for the base and the rate of divergence for the exponent. We can indeed obtain any result between $[0,infty)$ depending on that. That's why we define that an indeterminate form.



For example as $x to infty$




  • $forall ain mathbb{R}quad left(1+frac axright)^xto e^a$


and as a consequence




  • $left(1+frac axright)^{x^2}to infty quad a>0$

  • $left(1+frac axright)^{x^2}to 1 quad a=0$

  • $left(1+frac axright)^{x^2}to 0 quad a<0$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    For the sake of simplicity, I will show you a little bit different limit, but It's quite helpful in understanding that your limit is not $1$:
    $$x_n:=left(1+frac{a}{n}right)^n$$
    We can use Bernoulli's inequaluty to get that
    $$x_n geq 1+nfrac{a}{n}=1+a$$
    So the limit is at least $1+a$ (if it exists).

    And we don't need $n$ to be a natural number, it can be real as well, but it's much easier to prove the Bernoulli's inequality for natural exponents.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
      $endgroup$
      – gimusi
      Dec 9 '18 at 11:26












    • $begingroup$
      @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
      $endgroup$
      – Botond
      Dec 9 '18 at 11:34



















    2












    $begingroup$

    Notice that:



    $$lim_{x rightarrow infty} Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}$$



    By using the limit substitution rule ($xto infty$ means $x^2 to infty$)



    We know that this is a standard limit of the form:



    $$ lim_{prightarrow infty}left( 1+ frac{b}{p}right)^p = e^b$$



    So with this reasoning we get for $p =x^2$ and $b=-a^2$:
    $$ lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=e^{-a^2}$$



    Also see: https://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto%5Cinfty%7D%5CBigg(1-%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5CBigg)%5E%7Bx%5E2%7D





    As for your method of checking, I would generate a table (maybe in matlab or Mathematica) try for $a=2$, so we should get $e^{-4}approx 0.0183$.
    $$(1-4)^1=-3$$
    $$vdots$$
    $$(1-frac{4}{36})^{36} approx 0.01440$$
    $$(1-frac{4}{49})^{49} approx 0.01541$$
    For larger values the computer will start rounding and behaving weirdly, Wolfram alpa gives amazing fractions, but only up to some point. It will also be like "ehhh how about no" after some point and start rounding and truncating.
    https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B49%7D)%5E%7B49%7D





    Actually, wolfram alpha is really amazing, it can easily compute:
    $$ (1- frac{4}{1000})^{1000}=0.0181693095355$$
    with impeccable accuracy.



    https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B1000%7D)%5E%7B1000%7D



    At some point it WILL approximate:
    enter image description here






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      I agree with Rebellos on the first point. You can't ignore the exponent and take the limit of the inside first. This reasoning would give you e=1 (since e is a similarly defined limit). On the second point however, I disagree. Plugging in numbers is a completely valid way of estimating a limit. The problem is, calculators are not exact, they'll truncate the actual value to 1, and then raising 1, instead of 1+a, to the "huge number", (where a is some incredibly small number).






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        We are using that



        $$left(1-frac{a^2}{x^2}right)^{x^2}=e^{x^2 lnleft(1-frac{a^2}{x^2}right)}$$



        and $x^2 lnleft(1-frac{a^2}{x^2}right)$ is an indeterminate form $infty cdot 0$ and since the Taylor's expansion for $ln (1+t)$ as $tto 0$ is



        $$ln (1+t)=t-frac12t^2+frac13 t^3-frac14t^4+ldots$$



        the result follows.



        As an intuitive explanation of the result and for the indeterminate form $1^{infty}$, we need to consider that, in that case, while the base tends to $1^-$ the effect of the exponent is to make the quantity smaller towards zero. The final result is indeed a value in between that is $frac1{e^{a^2}}$.



        That's of course not always the case since all depends upon the rate of convergence to $1$ for the base and the rate of divergence for the exponent. We can indeed obtain any result between $[0,infty)$ depending on that. That's why we define that an indeterminate form.



        For example as $x to infty$




        • $forall ain mathbb{R}quad left(1+frac axright)^xto e^a$


        and as a consequence




        • $left(1+frac axright)^{x^2}to infty quad a>0$

        • $left(1+frac axright)^{x^2}to 1 quad a=0$

        • $left(1+frac axright)^{x^2}to 0 quad a<0$






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          We are using that



          $$left(1-frac{a^2}{x^2}right)^{x^2}=e^{x^2 lnleft(1-frac{a^2}{x^2}right)}$$



          and $x^2 lnleft(1-frac{a^2}{x^2}right)$ is an indeterminate form $infty cdot 0$ and since the Taylor's expansion for $ln (1+t)$ as $tto 0$ is



          $$ln (1+t)=t-frac12t^2+frac13 t^3-frac14t^4+ldots$$



          the result follows.



          As an intuitive explanation of the result and for the indeterminate form $1^{infty}$, we need to consider that, in that case, while the base tends to $1^-$ the effect of the exponent is to make the quantity smaller towards zero. The final result is indeed a value in between that is $frac1{e^{a^2}}$.



          That's of course not always the case since all depends upon the rate of convergence to $1$ for the base and the rate of divergence for the exponent. We can indeed obtain any result between $[0,infty)$ depending on that. That's why we define that an indeterminate form.



          For example as $x to infty$




          • $forall ain mathbb{R}quad left(1+frac axright)^xto e^a$


          and as a consequence




          • $left(1+frac axright)^{x^2}to infty quad a>0$

          • $left(1+frac axright)^{x^2}to 1 quad a=0$

          • $left(1+frac axright)^{x^2}to 0 quad a<0$






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            We are using that



            $$left(1-frac{a^2}{x^2}right)^{x^2}=e^{x^2 lnleft(1-frac{a^2}{x^2}right)}$$



            and $x^2 lnleft(1-frac{a^2}{x^2}right)$ is an indeterminate form $infty cdot 0$ and since the Taylor's expansion for $ln (1+t)$ as $tto 0$ is



            $$ln (1+t)=t-frac12t^2+frac13 t^3-frac14t^4+ldots$$



            the result follows.



            As an intuitive explanation of the result and for the indeterminate form $1^{infty}$, we need to consider that, in that case, while the base tends to $1^-$ the effect of the exponent is to make the quantity smaller towards zero. The final result is indeed a value in between that is $frac1{e^{a^2}}$.



            That's of course not always the case since all depends upon the rate of convergence to $1$ for the base and the rate of divergence for the exponent. We can indeed obtain any result between $[0,infty)$ depending on that. That's why we define that an indeterminate form.



            For example as $x to infty$




            • $forall ain mathbb{R}quad left(1+frac axright)^xto e^a$


            and as a consequence




            • $left(1+frac axright)^{x^2}to infty quad a>0$

            • $left(1+frac axright)^{x^2}to 1 quad a=0$

            • $left(1+frac axright)^{x^2}to 0 quad a<0$






            share|cite|improve this answer











            $endgroup$



            We are using that



            $$left(1-frac{a^2}{x^2}right)^{x^2}=e^{x^2 lnleft(1-frac{a^2}{x^2}right)}$$



            and $x^2 lnleft(1-frac{a^2}{x^2}right)$ is an indeterminate form $infty cdot 0$ and since the Taylor's expansion for $ln (1+t)$ as $tto 0$ is



            $$ln (1+t)=t-frac12t^2+frac13 t^3-frac14t^4+ldots$$



            the result follows.



            As an intuitive explanation of the result and for the indeterminate form $1^{infty}$, we need to consider that, in that case, while the base tends to $1^-$ the effect of the exponent is to make the quantity smaller towards zero. The final result is indeed a value in between that is $frac1{e^{a^2}}$.



            That's of course not always the case since all depends upon the rate of convergence to $1$ for the base and the rate of divergence for the exponent. We can indeed obtain any result between $[0,infty)$ depending on that. That's why we define that an indeterminate form.



            For example as $x to infty$




            • $forall ain mathbb{R}quad left(1+frac axright)^xto e^a$


            and as a consequence




            • $left(1+frac axright)^{x^2}to infty quad a>0$

            • $left(1+frac axright)^{x^2}to 1 quad a=0$

            • $left(1+frac axright)^{x^2}to 0 quad a<0$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 11:06

























            answered Dec 9 '18 at 10:21









            gimusigimusi

            92.9k94494




            92.9k94494























                2












                $begingroup$

                For the sake of simplicity, I will show you a little bit different limit, but It's quite helpful in understanding that your limit is not $1$:
                $$x_n:=left(1+frac{a}{n}right)^n$$
                We can use Bernoulli's inequaluty to get that
                $$x_n geq 1+nfrac{a}{n}=1+a$$
                So the limit is at least $1+a$ (if it exists).

                And we don't need $n$ to be a natural number, it can be real as well, but it's much easier to prove the Bernoulli's inequality for natural exponents.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
                  $endgroup$
                  – gimusi
                  Dec 9 '18 at 11:26












                • $begingroup$
                  @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
                  $endgroup$
                  – Botond
                  Dec 9 '18 at 11:34
















                2












                $begingroup$

                For the sake of simplicity, I will show you a little bit different limit, but It's quite helpful in understanding that your limit is not $1$:
                $$x_n:=left(1+frac{a}{n}right)^n$$
                We can use Bernoulli's inequaluty to get that
                $$x_n geq 1+nfrac{a}{n}=1+a$$
                So the limit is at least $1+a$ (if it exists).

                And we don't need $n$ to be a natural number, it can be real as well, but it's much easier to prove the Bernoulli's inequality for natural exponents.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
                  $endgroup$
                  – gimusi
                  Dec 9 '18 at 11:26












                • $begingroup$
                  @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
                  $endgroup$
                  – Botond
                  Dec 9 '18 at 11:34














                2












                2








                2





                $begingroup$

                For the sake of simplicity, I will show you a little bit different limit, but It's quite helpful in understanding that your limit is not $1$:
                $$x_n:=left(1+frac{a}{n}right)^n$$
                We can use Bernoulli's inequaluty to get that
                $$x_n geq 1+nfrac{a}{n}=1+a$$
                So the limit is at least $1+a$ (if it exists).

                And we don't need $n$ to be a natural number, it can be real as well, but it's much easier to prove the Bernoulli's inequality for natural exponents.






                share|cite|improve this answer









                $endgroup$



                For the sake of simplicity, I will show you a little bit different limit, but It's quite helpful in understanding that your limit is not $1$:
                $$x_n:=left(1+frac{a}{n}right)^n$$
                We can use Bernoulli's inequaluty to get that
                $$x_n geq 1+nfrac{a}{n}=1+a$$
                So the limit is at least $1+a$ (if it exists).

                And we don't need $n$ to be a natural number, it can be real as well, but it's much easier to prove the Bernoulli's inequality for natural exponents.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 '18 at 11:08









                BotondBotond

                5,6982732




                5,6982732












                • $begingroup$
                  That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
                  $endgroup$
                  – gimusi
                  Dec 9 '18 at 11:26












                • $begingroup$
                  @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
                  $endgroup$
                  – Botond
                  Dec 9 '18 at 11:34


















                • $begingroup$
                  That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
                  $endgroup$
                  – gimusi
                  Dec 9 '18 at 11:26












                • $begingroup$
                  @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
                  $endgroup$
                  – Botond
                  Dec 9 '18 at 11:34
















                $begingroup$
                That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
                $endgroup$
                – gimusi
                Dec 9 '18 at 11:26






                $begingroup$
                That's a valid argument only for $a>0$ if I'm not wrong. In that case we have $a<0$. Am I right or am I loosing something in your reasoning?
                $endgroup$
                – gimusi
                Dec 9 '18 at 11:26














                $begingroup$
                @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
                $endgroup$
                – Botond
                Dec 9 '18 at 11:34




                $begingroup$
                @gimusi Indeed, the inequality is true for $frac{a}{n} geq -1$ only, and useful when $a>0$. When $a<0$, we would need to get an upper bound for the sequence.
                $endgroup$
                – Botond
                Dec 9 '18 at 11:34











                2












                $begingroup$

                Notice that:



                $$lim_{x rightarrow infty} Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}$$



                By using the limit substitution rule ($xto infty$ means $x^2 to infty$)



                We know that this is a standard limit of the form:



                $$ lim_{prightarrow infty}left( 1+ frac{b}{p}right)^p = e^b$$



                So with this reasoning we get for $p =x^2$ and $b=-a^2$:
                $$ lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=e^{-a^2}$$



                Also see: https://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto%5Cinfty%7D%5CBigg(1-%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5CBigg)%5E%7Bx%5E2%7D





                As for your method of checking, I would generate a table (maybe in matlab or Mathematica) try for $a=2$, so we should get $e^{-4}approx 0.0183$.
                $$(1-4)^1=-3$$
                $$vdots$$
                $$(1-frac{4}{36})^{36} approx 0.01440$$
                $$(1-frac{4}{49})^{49} approx 0.01541$$
                For larger values the computer will start rounding and behaving weirdly, Wolfram alpa gives amazing fractions, but only up to some point. It will also be like "ehhh how about no" after some point and start rounding and truncating.
                https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B49%7D)%5E%7B49%7D





                Actually, wolfram alpha is really amazing, it can easily compute:
                $$ (1- frac{4}{1000})^{1000}=0.0181693095355$$
                with impeccable accuracy.



                https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B1000%7D)%5E%7B1000%7D



                At some point it WILL approximate:
                enter image description here






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  Notice that:



                  $$lim_{x rightarrow infty} Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}$$



                  By using the limit substitution rule ($xto infty$ means $x^2 to infty$)



                  We know that this is a standard limit of the form:



                  $$ lim_{prightarrow infty}left( 1+ frac{b}{p}right)^p = e^b$$



                  So with this reasoning we get for $p =x^2$ and $b=-a^2$:
                  $$ lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=e^{-a^2}$$



                  Also see: https://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto%5Cinfty%7D%5CBigg(1-%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5CBigg)%5E%7Bx%5E2%7D





                  As for your method of checking, I would generate a table (maybe in matlab or Mathematica) try for $a=2$, so we should get $e^{-4}approx 0.0183$.
                  $$(1-4)^1=-3$$
                  $$vdots$$
                  $$(1-frac{4}{36})^{36} approx 0.01440$$
                  $$(1-frac{4}{49})^{49} approx 0.01541$$
                  For larger values the computer will start rounding and behaving weirdly, Wolfram alpa gives amazing fractions, but only up to some point. It will also be like "ehhh how about no" after some point and start rounding and truncating.
                  https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B49%7D)%5E%7B49%7D





                  Actually, wolfram alpha is really amazing, it can easily compute:
                  $$ (1- frac{4}{1000})^{1000}=0.0181693095355$$
                  with impeccable accuracy.



                  https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B1000%7D)%5E%7B1000%7D



                  At some point it WILL approximate:
                  enter image description here






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Notice that:



                    $$lim_{x rightarrow infty} Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}$$



                    By using the limit substitution rule ($xto infty$ means $x^2 to infty$)



                    We know that this is a standard limit of the form:



                    $$ lim_{prightarrow infty}left( 1+ frac{b}{p}right)^p = e^b$$



                    So with this reasoning we get for $p =x^2$ and $b=-a^2$:
                    $$ lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=e^{-a^2}$$



                    Also see: https://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto%5Cinfty%7D%5CBigg(1-%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5CBigg)%5E%7Bx%5E2%7D





                    As for your method of checking, I would generate a table (maybe in matlab or Mathematica) try for $a=2$, so we should get $e^{-4}approx 0.0183$.
                    $$(1-4)^1=-3$$
                    $$vdots$$
                    $$(1-frac{4}{36})^{36} approx 0.01440$$
                    $$(1-frac{4}{49})^{49} approx 0.01541$$
                    For larger values the computer will start rounding and behaving weirdly, Wolfram alpa gives amazing fractions, but only up to some point. It will also be like "ehhh how about no" after some point and start rounding and truncating.
                    https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B49%7D)%5E%7B49%7D





                    Actually, wolfram alpha is really amazing, it can easily compute:
                    $$ (1- frac{4}{1000})^{1000}=0.0181693095355$$
                    with impeccable accuracy.



                    https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B1000%7D)%5E%7B1000%7D



                    At some point it WILL approximate:
                    enter image description here






                    share|cite|improve this answer











                    $endgroup$



                    Notice that:



                    $$lim_{x rightarrow infty} Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}$$



                    By using the limit substitution rule ($xto infty$ means $x^2 to infty$)



                    We know that this is a standard limit of the form:



                    $$ lim_{prightarrow infty}left( 1+ frac{b}{p}right)^p = e^b$$



                    So with this reasoning we get for $p =x^2$ and $b=-a^2$:
                    $$ lim_{x^2toinfty}Bigg(1-frac{a^2}{x^2}Bigg)^{x^2}=e^{-a^2}$$



                    Also see: https://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto%5Cinfty%7D%5CBigg(1-%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5CBigg)%5E%7Bx%5E2%7D





                    As for your method of checking, I would generate a table (maybe in matlab or Mathematica) try for $a=2$, so we should get $e^{-4}approx 0.0183$.
                    $$(1-4)^1=-3$$
                    $$vdots$$
                    $$(1-frac{4}{36})^{36} approx 0.01440$$
                    $$(1-frac{4}{49})^{49} approx 0.01541$$
                    For larger values the computer will start rounding and behaving weirdly, Wolfram alpa gives amazing fractions, but only up to some point. It will also be like "ehhh how about no" after some point and start rounding and truncating.
                    https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B49%7D)%5E%7B49%7D





                    Actually, wolfram alpha is really amazing, it can easily compute:
                    $$ (1- frac{4}{1000})^{1000}=0.0181693095355$$
                    with impeccable accuracy.



                    https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B1000%7D)%5E%7B1000%7D



                    At some point it WILL approximate:
                    enter image description here







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 9 '18 at 11:29

























                    answered Dec 9 '18 at 10:52









                    Wesley StrikWesley Strik

                    1,665423




                    1,665423























                        1












                        $begingroup$

                        I agree with Rebellos on the first point. You can't ignore the exponent and take the limit of the inside first. This reasoning would give you e=1 (since e is a similarly defined limit). On the second point however, I disagree. Plugging in numbers is a completely valid way of estimating a limit. The problem is, calculators are not exact, they'll truncate the actual value to 1, and then raising 1, instead of 1+a, to the "huge number", (where a is some incredibly small number).






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          I agree with Rebellos on the first point. You can't ignore the exponent and take the limit of the inside first. This reasoning would give you e=1 (since e is a similarly defined limit). On the second point however, I disagree. Plugging in numbers is a completely valid way of estimating a limit. The problem is, calculators are not exact, they'll truncate the actual value to 1, and then raising 1, instead of 1+a, to the "huge number", (where a is some incredibly small number).






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            I agree with Rebellos on the first point. You can't ignore the exponent and take the limit of the inside first. This reasoning would give you e=1 (since e is a similarly defined limit). On the second point however, I disagree. Plugging in numbers is a completely valid way of estimating a limit. The problem is, calculators are not exact, they'll truncate the actual value to 1, and then raising 1, instead of 1+a, to the "huge number", (where a is some incredibly small number).






                            share|cite|improve this answer









                            $endgroup$



                            I agree with Rebellos on the first point. You can't ignore the exponent and take the limit of the inside first. This reasoning would give you e=1 (since e is a similarly defined limit). On the second point however, I disagree. Plugging in numbers is a completely valid way of estimating a limit. The problem is, calculators are not exact, they'll truncate the actual value to 1, and then raising 1, instead of 1+a, to the "huge number", (where a is some incredibly small number).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 9 '18 at 10:24









                            BelowAverageIntelligenceBelowAverageIntelligence

                            5021213




                            5021213






























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