How to find $lim_{x to infty} (1+sinh(x^{-2}))^{(x^2)}$ [on hold]












-2














I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.










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put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
    – Mark Viola
    Jan 3 at 21:38










  • Yes, sorry, we use sh for sinh
    – MilO
    Jan 3 at 21:42










  • Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
    – Sauhard Sharma
    Jan 3 at 21:43










  • You don't want answers that use L'Hopital's rule?
    – 0x539
    Jan 3 at 21:53
















-2














I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.










share|cite|improve this question









New contributor




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











put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
    – Mark Viola
    Jan 3 at 21:38










  • Yes, sorry, we use sh for sinh
    – MilO
    Jan 3 at 21:42










  • Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
    – Sauhard Sharma
    Jan 3 at 21:43










  • You don't want answers that use L'Hopital's rule?
    – 0x539
    Jan 3 at 21:53














-2












-2








-2







I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.










share|cite|improve this question









New contributor




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











I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.







calculus limits limits-without-lhopital






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











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edited Jan 3 at 21:50









0x539

1,047316




1,047316






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asked Jan 3 at 21:31









MilO

61




61




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




put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
    – Mark Viola
    Jan 3 at 21:38










  • Yes, sorry, we use sh for sinh
    – MilO
    Jan 3 at 21:42










  • Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
    – Sauhard Sharma
    Jan 3 at 21:43










  • You don't want answers that use L'Hopital's rule?
    – 0x539
    Jan 3 at 21:53














  • 1




    What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
    – Mark Viola
    Jan 3 at 21:38










  • Yes, sorry, we use sh for sinh
    – MilO
    Jan 3 at 21:42










  • Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
    – Sauhard Sharma
    Jan 3 at 21:43










  • You don't want answers that use L'Hopital's rule?
    – 0x539
    Jan 3 at 21:53








1




1




What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38




What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38












Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42




Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42












Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43




Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43












You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53




You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53










2 Answers
2






active

oldest

votes


















1














The usual idea is to compute
$$
l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
$$

and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
$$
lim_{tto0^+}frac{log(1+sinh t)}{t}
$$






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    0














    This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
    $$
    lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
    $$



    RHS is a very famous limit.






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      The usual idea is to compute
      $$
      l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
      $$

      and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
      $$
      lim_{tto0^+}frac{log(1+sinh t)}{t}
      $$






      share|cite|improve this answer


























        1














        The usual idea is to compute
        $$
        l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
        $$

        and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
        $$
        lim_{tto0^+}frac{log(1+sinh t)}{t}
        $$






        share|cite|improve this answer
























          1












          1








          1






          The usual idea is to compute
          $$
          l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
          $$

          and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
          $$
          lim_{tto0^+}frac{log(1+sinh t)}{t}
          $$






          share|cite|improve this answer












          The usual idea is to compute
          $$
          l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
          $$

          and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
          $$
          lim_{tto0^+}frac{log(1+sinh t)}{t}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 21:53









          egreg

          179k1484201




          179k1484201























              0














              This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
              $$
              lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
              $$



              RHS is a very famous limit.






              share|cite|improve this answer


























                0














                This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
                $$
                lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
                $$



                RHS is a very famous limit.






                share|cite|improve this answer
























                  0












                  0








                  0






                  This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
                  $$
                  lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
                  $$



                  RHS is a very famous limit.






                  share|cite|improve this answer












                  This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
                  $$
                  lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
                  $$



                  RHS is a very famous limit.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 22:22









                  roman

                  1,98121221




                  1,98121221















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