What is the limit of the following expression?











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I've been thinking about and trying to solve the following limit that I just feel lost by now. I always get an indeterminate form. I don't know what else to try. In the picture is just one way that I tried to do it, again resulting in an indeterminate:
enter image description here



Can you help?



$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$










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  • Central limit theorem? Where?
    – Federico
    Nov 21 at 19:48










  • Do you know anything about $left(a+frac1nright)^n?$
    – Federico
    Nov 21 at 19:49










  • Bogdan, this looks like a good question. Would you mind putting it in MathJax? I would if I could, but am currently away from my computer.
    – Jack Moody
    Nov 21 at 19:49










  • @Federico Yes, but I don't see how to apply it here.
    – Bogdan Vlad
    Nov 21 at 20:20






  • 1




    @JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
    – Bogdan Vlad
    Nov 21 at 22:00















up vote
1
down vote

favorite












I've been thinking about and trying to solve the following limit that I just feel lost by now. I always get an indeterminate form. I don't know what else to try. In the picture is just one way that I tried to do it, again resulting in an indeterminate:
enter image description here



Can you help?



$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$










share|cite|improve this question
























  • Central limit theorem? Where?
    – Federico
    Nov 21 at 19:48










  • Do you know anything about $left(a+frac1nright)^n?$
    – Federico
    Nov 21 at 19:49










  • Bogdan, this looks like a good question. Would you mind putting it in MathJax? I would if I could, but am currently away from my computer.
    – Jack Moody
    Nov 21 at 19:49










  • @Federico Yes, but I don't see how to apply it here.
    – Bogdan Vlad
    Nov 21 at 20:20






  • 1




    @JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
    – Bogdan Vlad
    Nov 21 at 22:00













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I've been thinking about and trying to solve the following limit that I just feel lost by now. I always get an indeterminate form. I don't know what else to try. In the picture is just one way that I tried to do it, again resulting in an indeterminate:
enter image description here



Can you help?



$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$










share|cite|improve this question















I've been thinking about and trying to solve the following limit that I just feel lost by now. I always get an indeterminate form. I don't know what else to try. In the picture is just one way that I tried to do it, again resulting in an indeterminate:
enter image description here



Can you help?



$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$







limits central-limit-theorem indeterminate-forms






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 21:59

























asked Nov 21 at 19:44









Bogdan Vlad

495




495












  • Central limit theorem? Where?
    – Federico
    Nov 21 at 19:48










  • Do you know anything about $left(a+frac1nright)^n?$
    – Federico
    Nov 21 at 19:49










  • Bogdan, this looks like a good question. Would you mind putting it in MathJax? I would if I could, but am currently away from my computer.
    – Jack Moody
    Nov 21 at 19:49










  • @Federico Yes, but I don't see how to apply it here.
    – Bogdan Vlad
    Nov 21 at 20:20






  • 1




    @JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
    – Bogdan Vlad
    Nov 21 at 22:00


















  • Central limit theorem? Where?
    – Federico
    Nov 21 at 19:48










  • Do you know anything about $left(a+frac1nright)^n?$
    – Federico
    Nov 21 at 19:49










  • Bogdan, this looks like a good question. Would you mind putting it in MathJax? I would if I could, but am currently away from my computer.
    – Jack Moody
    Nov 21 at 19:49










  • @Federico Yes, but I don't see how to apply it here.
    – Bogdan Vlad
    Nov 21 at 20:20






  • 1




    @JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
    – Bogdan Vlad
    Nov 21 at 22:00
















Central limit theorem? Where?
– Federico
Nov 21 at 19:48




Central limit theorem? Where?
– Federico
Nov 21 at 19:48












Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 at 19:49




Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 at 19:49












Bogdan, this looks like a good question. Would you mind putting it in MathJax? I would if I could, but am currently away from my computer.
– Jack Moody
Nov 21 at 19:49




Bogdan, this looks like a good question. Would you mind putting it in MathJax? I would if I could, but am currently away from my computer.
– Jack Moody
Nov 21 at 19:49












@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 at 20:20




@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 at 20:20




1




1




@JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
– Bogdan Vlad
Nov 21 at 22:00




@JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
– Bogdan Vlad
Nov 21 at 22:00










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Let $x=dfrac{t}{3}$ then your limit will be
$$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$






share|cite|improve this answer





















  • I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
    – Bogdan Vlad
    Nov 21 at 20:24










  • Nevermind, I got it (with your idea). Thanks a lot!
    – Bogdan Vlad
    Nov 21 at 20:59












  • You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
    – Nosrati
    Nov 22 at 5:23




















up vote
0
down vote













My suggestion was to notice that
$$
left(a+frac1nright)^n = left(a+frac{a}{an}right)^n
= a^nleft(1+frac1{an}right)^n sim a^ne^{1/a}.
$$



You can compute your limit as
$$
left(3+frac1nright)^{-3n}3^{3n}
= 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
to e^{-1}.
$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Let $x=dfrac{t}{3}$ then your limit will be
    $$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$






    share|cite|improve this answer





















    • I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
      – Bogdan Vlad
      Nov 21 at 20:24










    • Nevermind, I got it (with your idea). Thanks a lot!
      – Bogdan Vlad
      Nov 21 at 20:59












    • You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
      – Nosrati
      Nov 22 at 5:23

















    up vote
    1
    down vote



    accepted










    Let $x=dfrac{t}{3}$ then your limit will be
    $$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$






    share|cite|improve this answer





















    • I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
      – Bogdan Vlad
      Nov 21 at 20:24










    • Nevermind, I got it (with your idea). Thanks a lot!
      – Bogdan Vlad
      Nov 21 at 20:59












    • You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
      – Nosrati
      Nov 22 at 5:23















    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Let $x=dfrac{t}{3}$ then your limit will be
    $$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$






    share|cite|improve this answer












    Let $x=dfrac{t}{3}$ then your limit will be
    $$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 at 19:49









    Nosrati

    26.3k62353




    26.3k62353












    • I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
      – Bogdan Vlad
      Nov 21 at 20:24










    • Nevermind, I got it (with your idea). Thanks a lot!
      – Bogdan Vlad
      Nov 21 at 20:59












    • You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
      – Nosrati
      Nov 22 at 5:23




















    • I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
      – Bogdan Vlad
      Nov 21 at 20:24










    • Nevermind, I got it (with your idea). Thanks a lot!
      – Bogdan Vlad
      Nov 21 at 20:59












    • You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
      – Nosrati
      Nov 22 at 5:23


















    I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
    – Bogdan Vlad
    Nov 21 at 20:24




    I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
    – Bogdan Vlad
    Nov 21 at 20:24












    Nevermind, I got it (with your idea). Thanks a lot!
    – Bogdan Vlad
    Nov 21 at 20:59






    Nevermind, I got it (with your idea). Thanks a lot!
    – Bogdan Vlad
    Nov 21 at 20:59














    You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
    – Nosrati
    Nov 22 at 5:23






    You should replace $x=dfrac{t}{3}$ at the first limit $$(3+dfrac{1}{frac{t}{3}})^{-3frac{t}{3}}=(3+dfrac{3}{t})^{-t}=3^{-t}(1+dfrac{1}{t})^{-t}$$also $$27^x=(3^3)^dfrac{t}{3}=3^t$$
    – Nosrati
    Nov 22 at 5:23












    up vote
    0
    down vote













    My suggestion was to notice that
    $$
    left(a+frac1nright)^n = left(a+frac{a}{an}right)^n
    = a^nleft(1+frac1{an}right)^n sim a^ne^{1/a}.
    $$



    You can compute your limit as
    $$
    left(3+frac1nright)^{-3n}3^{3n}
    = 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
    to e^{-1}.
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      My suggestion was to notice that
      $$
      left(a+frac1nright)^n = left(a+frac{a}{an}right)^n
      = a^nleft(1+frac1{an}right)^n sim a^ne^{1/a}.
      $$



      You can compute your limit as
      $$
      left(3+frac1nright)^{-3n}3^{3n}
      = 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
      to e^{-1}.
      $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        My suggestion was to notice that
        $$
        left(a+frac1nright)^n = left(a+frac{a}{an}right)^n
        = a^nleft(1+frac1{an}right)^n sim a^ne^{1/a}.
        $$



        You can compute your limit as
        $$
        left(3+frac1nright)^{-3n}3^{3n}
        = 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
        to e^{-1}.
        $$






        share|cite|improve this answer












        My suggestion was to notice that
        $$
        left(a+frac1nright)^n = left(a+frac{a}{an}right)^n
        = a^nleft(1+frac1{an}right)^n sim a^ne^{1/a}.
        $$



        You can compute your limit as
        $$
        left(3+frac1nright)^{-3n}3^{3n}
        = 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
        to e^{-1}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 17:38









        Federico

        4,138512




        4,138512






























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