Proving a second order special limit without derivatives












0












$begingroup$


The special limit
$$
lim_{x to 0} frac{e^x-x-1}{x^2}=frac 1 2
$$
can be proved by Taylor expansion or with L'Hôpital's rule. Is it possoble to prove it without using derivatives?






real-analysis limits limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is there a reason that neither of these methods can be used?
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:54






  • 1




    $begingroup$
    If we define $e^x$ as $sum_{ngeq 0}frac{x^n}{n!}$ that is trivial.
    $endgroup$
    – Jack D'Aurizio
    Dec 2 '18 at 19:07










  • $begingroup$
    See math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 8:45
















0












$begingroup$


The special limit
$$
lim_{x to 0} frac{e^x-x-1}{x^2}=frac 1 2
$$
can be proved by Taylor expansion or with L'Hôpital's rule. Is it possoble to prove it without using derivatives?






real-analysis limits limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is there a reason that neither of these methods can be used?
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:54






  • 1




    $begingroup$
    If we define $e^x$ as $sum_{ngeq 0}frac{x^n}{n!}$ that is trivial.
    $endgroup$
    – Jack D'Aurizio
    Dec 2 '18 at 19:07










  • $begingroup$
    See math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 8:45














0












0








0





$begingroup$


The special limit
$$
lim_{x to 0} frac{e^x-x-1}{x^2}=frac 1 2
$$
can be proved by Taylor expansion or with L'Hôpital's rule. Is it possoble to prove it without using derivatives?






real-analysis limits limits-without-lhopital







share|cite|improve this question











$endgroup$




The special limit
$$
lim_{x to 0} frac{e^x-x-1}{x^2}=frac 1 2
$$
can be proved by Taylor expansion or with L'Hôpital's rule. Is it possoble to prove it without using derivatives?






real-analysis limits limits-without-lhopital




real-analysis limits limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 19:38









gimusi

92.9k94494




92.9k94494










asked Dec 2 '18 at 18:52









Marco DisceMarco Disce

1,3961216




1,3961216








  • 2




    $begingroup$
    Is there a reason that neither of these methods can be used?
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:54






  • 1




    $begingroup$
    If we define $e^x$ as $sum_{ngeq 0}frac{x^n}{n!}$ that is trivial.
    $endgroup$
    – Jack D'Aurizio
    Dec 2 '18 at 19:07










  • $begingroup$
    See math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 8:45














  • 2




    $begingroup$
    Is there a reason that neither of these methods can be used?
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:54






  • 1




    $begingroup$
    If we define $e^x$ as $sum_{ngeq 0}frac{x^n}{n!}$ that is trivial.
    $endgroup$
    – Jack D'Aurizio
    Dec 2 '18 at 19:07










  • $begingroup$
    See math.stackexchange.com/questions/387333/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 8:45








2




2




$begingroup$
Is there a reason that neither of these methods can be used?
$endgroup$
– Henry Lee
Dec 2 '18 at 18:54




$begingroup$
Is there a reason that neither of these methods can be used?
$endgroup$
– Henry Lee
Dec 2 '18 at 18:54




1




1




$begingroup$
If we define $e^x$ as $sum_{ngeq 0}frac{x^n}{n!}$ that is trivial.
$endgroup$
– Jack D'Aurizio
Dec 2 '18 at 19:07




$begingroup$
If we define $e^x$ as $sum_{ngeq 0}frac{x^n}{n!}$ that is trivial.
$endgroup$
– Jack D'Aurizio
Dec 2 '18 at 19:07












$begingroup$
See math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 8:45




$begingroup$
See math.stackexchange.com/questions/387333/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 8:45










3 Answers
3






active

oldest

votes


















2












$begingroup$

$$frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y)e^{xy},dy $$
hence by the dominated convergence theorem
$$ lim_{xto 0}frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y),dy = frac{1}{2}.$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We have by $x=2y$



    $$frac{e^x-x-1}{x^2}=frac{e^{2y}-2y-1}{4y^2}=frac{(e^y-1)^2+2e^y-2y-2}{4y^2}=frac14left(frac{e^y-1}{y}right)^2+frac12frac{e^y-y-1}{y^2}$$



    therefore assuming that the limit exists we have



    $$L=frac14+frac12L implies L=frac12$$



    Refer also to




    • Are all limits solvable without L'Hôpital Rule or Series Expansion






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
      $endgroup$
      – Martin R
      Dec 2 '18 at 21:25












    • $begingroup$
      @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
      $endgroup$
      – gimusi
      Dec 2 '18 at 21:38



















    0












    $begingroup$

    Let $$f(x)=e^{sqrt{x}}-sqrt{x}$$



    Then your limit is



    $$lim_{Xto0^+}frac{f(X)-f(0)}{X-0}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      In the limit expression, you mean $f(x^2)$.
      $endgroup$
      – Anurag A
      Dec 2 '18 at 19:01












    • $begingroup$
      @AnuragA No. {}{}{}{}{}{}.
      $endgroup$
      – hamam_Abdallah
      Dec 2 '18 at 19:01










    • $begingroup$
      is that by derivative?
      $endgroup$
      – gimusi
      Dec 2 '18 at 20:06











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $$frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y)e^{xy},dy $$
    hence by the dominated convergence theorem
    $$ lim_{xto 0}frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y),dy = frac{1}{2}.$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $$frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y)e^{xy},dy $$
      hence by the dominated convergence theorem
      $$ lim_{xto 0}frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y),dy = frac{1}{2}.$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $$frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y)e^{xy},dy $$
        hence by the dominated convergence theorem
        $$ lim_{xto 0}frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y),dy = frac{1}{2}.$$






        share|cite|improve this answer









        $endgroup$



        $$frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y)e^{xy},dy $$
        hence by the dominated convergence theorem
        $$ lim_{xto 0}frac{e^x-1-x}{x^2}=int_{0}^{1}(1-y),dy = frac{1}{2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 19:10









        Jack D'AurizioJack D'Aurizio

        288k33280660




        288k33280660























            1












            $begingroup$

            We have by $x=2y$



            $$frac{e^x-x-1}{x^2}=frac{e^{2y}-2y-1}{4y^2}=frac{(e^y-1)^2+2e^y-2y-2}{4y^2}=frac14left(frac{e^y-1}{y}right)^2+frac12frac{e^y-y-1}{y^2}$$



            therefore assuming that the limit exists we have



            $$L=frac14+frac12L implies L=frac12$$



            Refer also to




            • Are all limits solvable without L'Hôpital Rule or Series Expansion






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
              $endgroup$
              – Martin R
              Dec 2 '18 at 21:25












            • $begingroup$
              @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
              $endgroup$
              – gimusi
              Dec 2 '18 at 21:38
















            1












            $begingroup$

            We have by $x=2y$



            $$frac{e^x-x-1}{x^2}=frac{e^{2y}-2y-1}{4y^2}=frac{(e^y-1)^2+2e^y-2y-2}{4y^2}=frac14left(frac{e^y-1}{y}right)^2+frac12frac{e^y-y-1}{y^2}$$



            therefore assuming that the limit exists we have



            $$L=frac14+frac12L implies L=frac12$$



            Refer also to




            • Are all limits solvable without L'Hôpital Rule or Series Expansion






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
              $endgroup$
              – Martin R
              Dec 2 '18 at 21:25












            • $begingroup$
              @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
              $endgroup$
              – gimusi
              Dec 2 '18 at 21:38














            1












            1








            1





            $begingroup$

            We have by $x=2y$



            $$frac{e^x-x-1}{x^2}=frac{e^{2y}-2y-1}{4y^2}=frac{(e^y-1)^2+2e^y-2y-2}{4y^2}=frac14left(frac{e^y-1}{y}right)^2+frac12frac{e^y-y-1}{y^2}$$



            therefore assuming that the limit exists we have



            $$L=frac14+frac12L implies L=frac12$$



            Refer also to




            • Are all limits solvable without L'Hôpital Rule or Series Expansion






            share|cite|improve this answer









            $endgroup$



            We have by $x=2y$



            $$frac{e^x-x-1}{x^2}=frac{e^{2y}-2y-1}{4y^2}=frac{(e^y-1)^2+2e^y-2y-2}{4y^2}=frac14left(frac{e^y-1}{y}right)^2+frac12frac{e^y-y-1}{y^2}$$



            therefore assuming that the limit exists we have



            $$L=frac14+frac12L implies L=frac12$$



            Refer also to




            • Are all limits solvable without L'Hôpital Rule or Series Expansion







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 '18 at 19:11









            gimusigimusi

            92.9k94494




            92.9k94494












            • $begingroup$
              Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
              $endgroup$
              – Martin R
              Dec 2 '18 at 21:25












            • $begingroup$
              @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
              $endgroup$
              – gimusi
              Dec 2 '18 at 21:38


















            • $begingroup$
              Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
              $endgroup$
              – Martin R
              Dec 2 '18 at 21:25












            • $begingroup$
              @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
              $endgroup$
              – gimusi
              Dec 2 '18 at 21:38
















            $begingroup$
            Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
            $endgroup$
            – Martin R
            Dec 2 '18 at 21:25






            $begingroup$
            Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y to 1$ is obtained.
            $endgroup$
            – Martin R
            Dec 2 '18 at 21:25














            $begingroup$
            @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
            $endgroup$
            – gimusi
            Dec 2 '18 at 21:38




            $begingroup$
            @MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks
            $endgroup$
            – gimusi
            Dec 2 '18 at 21:38











            0












            $begingroup$

            Let $$f(x)=e^{sqrt{x}}-sqrt{x}$$



            Then your limit is



            $$lim_{Xto0^+}frac{f(X)-f(0)}{X-0}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              In the limit expression, you mean $f(x^2)$.
              $endgroup$
              – Anurag A
              Dec 2 '18 at 19:01












            • $begingroup$
              @AnuragA No. {}{}{}{}{}{}.
              $endgroup$
              – hamam_Abdallah
              Dec 2 '18 at 19:01










            • $begingroup$
              is that by derivative?
              $endgroup$
              – gimusi
              Dec 2 '18 at 20:06
















            0












            $begingroup$

            Let $$f(x)=e^{sqrt{x}}-sqrt{x}$$



            Then your limit is



            $$lim_{Xto0^+}frac{f(X)-f(0)}{X-0}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              In the limit expression, you mean $f(x^2)$.
              $endgroup$
              – Anurag A
              Dec 2 '18 at 19:01












            • $begingroup$
              @AnuragA No. {}{}{}{}{}{}.
              $endgroup$
              – hamam_Abdallah
              Dec 2 '18 at 19:01










            • $begingroup$
              is that by derivative?
              $endgroup$
              – gimusi
              Dec 2 '18 at 20:06














            0












            0








            0





            $begingroup$

            Let $$f(x)=e^{sqrt{x}}-sqrt{x}$$



            Then your limit is



            $$lim_{Xto0^+}frac{f(X)-f(0)}{X-0}$$






            share|cite|improve this answer









            $endgroup$



            Let $$f(x)=e^{sqrt{x}}-sqrt{x}$$



            Then your limit is



            $$lim_{Xto0^+}frac{f(X)-f(0)}{X-0}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 '18 at 18:59









            hamam_Abdallahhamam_Abdallah

            38k21634




            38k21634












            • $begingroup$
              In the limit expression, you mean $f(x^2)$.
              $endgroup$
              – Anurag A
              Dec 2 '18 at 19:01












            • $begingroup$
              @AnuragA No. {}{}{}{}{}{}.
              $endgroup$
              – hamam_Abdallah
              Dec 2 '18 at 19:01










            • $begingroup$
              is that by derivative?
              $endgroup$
              – gimusi
              Dec 2 '18 at 20:06


















            • $begingroup$
              In the limit expression, you mean $f(x^2)$.
              $endgroup$
              – Anurag A
              Dec 2 '18 at 19:01












            • $begingroup$
              @AnuragA No. {}{}{}{}{}{}.
              $endgroup$
              – hamam_Abdallah
              Dec 2 '18 at 19:01










            • $begingroup$
              is that by derivative?
              $endgroup$
              – gimusi
              Dec 2 '18 at 20:06
















            $begingroup$
            In the limit expression, you mean $f(x^2)$.
            $endgroup$
            – Anurag A
            Dec 2 '18 at 19:01






            $begingroup$
            In the limit expression, you mean $f(x^2)$.
            $endgroup$
            – Anurag A
            Dec 2 '18 at 19:01














            $begingroup$
            @AnuragA No. {}{}{}{}{}{}.
            $endgroup$
            – hamam_Abdallah
            Dec 2 '18 at 19:01




            $begingroup$
            @AnuragA No. {}{}{}{}{}{}.
            $endgroup$
            – hamam_Abdallah
            Dec 2 '18 at 19:01












            $begingroup$
            is that by derivative?
            $endgroup$
            – gimusi
            Dec 2 '18 at 20:06




            $begingroup$
            is that by derivative?
            $endgroup$
            – gimusi
            Dec 2 '18 at 20:06


















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