Indeterminate form and L'Hospital rule












1












$begingroup$



$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58
















1












$begingroup$



$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58














1












1








1





$begingroup$



$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus







share|cite|improve this question











$endgroup$





$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 11:06









Namaste

1




1










asked Dec 16 '18 at 10:55









Jobiar HossainJobiar Hossain

414




414












  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58


















  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58
















$begingroup$
$infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
$endgroup$
– Viktor Glombik
Dec 16 '18 at 10:58




$begingroup$
$infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
$endgroup$
– Viktor Glombik
Dec 16 '18 at 10:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



As an alternative note that eventually $e^xge x^3$ and then



$$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



We can prove that also by induction, that is $e^n ge n^3$




  • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

  • induction step: assume true $e^k ge k^3$ for some $k$ then


$$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



and the latter is true indeed



$$2cdot k^3ge (k+1)^3$$



$$k(k^2-3k-3)ge 1, quad kge 4$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Note that $infty$ is not an intermediate form. Simply put :



    $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



    Indeterminate forms are typically considered to be :



    $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I guess you intended to write "indeterminate", not "intermediate"
      $endgroup$
      – jjagmath
      Dec 16 '18 at 11:37










    • $begingroup$
      @jjagmath Thanks!
      $endgroup$
      – Rebellos
      Dec 16 '18 at 11:37











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



    As an alternative note that eventually $e^xge x^3$ and then



    $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



    We can prove that also by induction, that is $e^n ge n^3$




    • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

    • induction step: assume true $e^k ge k^3$ for some $k$ then


    $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



    and the latter is true indeed



    $$2cdot k^3ge (k+1)^3$$



    $$k(k^2-3k-3)ge 1, quad kge 4$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



      As an alternative note that eventually $e^xge x^3$ and then



      $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



      We can prove that also by induction, that is $e^n ge n^3$




      • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

      • induction step: assume true $e^k ge k^3$ for some $k$ then


      $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



      and the latter is true indeed



      $$2cdot k^3ge (k+1)^3$$



      $$k(k^2-3k-3)ge 1, quad kge 4$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



        As an alternative note that eventually $e^xge x^3$ and then



        $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



        We can prove that also by induction, that is $e^n ge n^3$




        • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

        • induction step: assume true $e^k ge k^3$ for some $k$ then


        $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



        and the latter is true indeed



        $$2cdot k^3ge (k+1)^3$$



        $$k(k^2-3k-3)ge 1, quad kge 4$$






        share|cite|improve this answer











        $endgroup$



        Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



        As an alternative note that eventually $e^xge x^3$ and then



        $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



        We can prove that also by induction, that is $e^n ge n^3$




        • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

        • induction step: assume true $e^k ge k^3$ for some $k$ then


        $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



        and the latter is true indeed



        $$2cdot k^3ge (k+1)^3$$



        $$k(k^2-3k-3)ge 1, quad kge 4$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 11:29

























        answered Dec 16 '18 at 10:57









        gimusigimusi

        93k84494




        93k84494























            1












            $begingroup$

            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37
















            1












            $begingroup$

            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37














            1












            1








            1





            $begingroup$

            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






            share|cite|improve this answer











            $endgroup$



            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 16 '18 at 11:37

























            answered Dec 16 '18 at 11:23









            RebellosRebellos

            15.3k31250




            15.3k31250












            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37


















            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37
















            $begingroup$
            I guess you intended to write "indeterminate", not "intermediate"
            $endgroup$
            – jjagmath
            Dec 16 '18 at 11:37




            $begingroup$
            I guess you intended to write "indeterminate", not "intermediate"
            $endgroup$
            – jjagmath
            Dec 16 '18 at 11:37












            $begingroup$
            @jjagmath Thanks!
            $endgroup$
            – Rebellos
            Dec 16 '18 at 11:37




            $begingroup$
            @jjagmath Thanks!
            $endgroup$
            – Rebellos
            Dec 16 '18 at 11:37


















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