Mixing
What is the limit of the following expression?
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:
Can you help?
$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$
limits central-limit-theorem indeterminate-forms
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
|
show 3 more comments
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:
Can you help?
$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$
limits central-limit-theorem indeterminate-forms
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
Central limit theorem? Where?
– Federico
Nov 21 '18 at 19:48
Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 '18 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 '18 at 19:49
@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 '18 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 '18 at 22:00
|
show 3 more comments
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:
Can you help?
$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$
limits central-limit-theorem indeterminate-forms
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
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:
Can you help?
$lim_{nto infty} left[ left(3+dfrac{1}{n} right)^{-3n} * 27^nright]$
limits central-limit-theorem indeterminate-forms
limits central-limit-theorem indeterminate-forms
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
edited Nov 21 '18 at 21:59
Bogdan Vlad
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
asked Nov 21 '18 at 19:44
Bogdan VladBogdan Vlad
495
495
Central limit theorem? Where?
– Federico
Nov 21 '18 at 19:48
Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 '18 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 '18 at 19:49
@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 '18 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 '18 at 22:00
|
show 3 more comments
Central limit theorem? Where?
– Federico
Nov 21 '18 at 19:48
Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 '18 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 '18 at 19:49
@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 '18 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 '18 at 22:00
Central limit theorem? Where?
– Federico
Nov 21 '18 at 19:48
Central limit theorem? Where?
– Federico
Nov 21 '18 at 19:48
Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 '18 at 19:49
Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 '18 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 '18 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 '18 at 19:49
@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 '18 at 20:20
@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 '18 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 '18 at 22:00
@JackMoody Oh, my bad. I didn't know. I put it at the end of the question.
– Bogdan Vlad
Nov 21 '18 at 22:00
|
show 3 more comments
2 Answers
2
active
oldest
votes
Let $x=dfrac{t}{3}$ then your limit will be
$$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
– Bogdan Vlad
Nov 21 '18 at 20:24
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 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 '18 at 5:23
add a comment |
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
$$
left(3+frac1nright)^{-3n}3^{3n}
= 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
to e^{-1}.
$$
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x=dfrac{t}{3}$ then your limit will be
$$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
– Bogdan Vlad
Nov 21 '18 at 20:24
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 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 '18 at 5:23
add a comment |
Let $x=dfrac{t}{3}$ then your limit will be
$$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
– Bogdan Vlad
Nov 21 '18 at 20:24
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 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 '18 at 5:23
add a comment |
Let $x=dfrac{t}{3}$ then your limit will be
$$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
Let $x=dfrac{t}{3}$ then your limit will be
$$lim_{ttoinfty}left(1+dfrac{1}{t}right)^{-t}=dfrac{1}{e}$$
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
answered Nov 21 '18 at 19:49
NosratiNosrati
26.5k62354
26.5k62354
I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
– Bogdan Vlad
Nov 21 '18 at 20:24
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 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 '18 at 5:23
add a comment |
I'm sorry but can you please explain it a bit further? Where and what should I replace with $t$ ?
– Bogdan Vlad
Nov 21 '18 at 20:24
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 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 '18 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 '18 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 '18 at 20:24
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 at 20:59
Nevermind, I got it (with your idea). Thanks a lot!
– Bogdan Vlad
Nov 21 '18 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 '18 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 '18 at 5:23
add a comment |
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
$$
left(3+frac1nright)^{-3n}3^{3n}
= 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
to e^{-1}.
$$
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
add a comment |
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
$$
left(3+frac1nright)^{-3n}3^{3n}
= 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
to e^{-1}.
$$
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
add a comment |
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
$$
left(3+frac1nright)^{-3n}3^{3n}
= 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
to e^{-1}.
$$
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
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
$$
left(3+frac1nright)^{-3n}3^{3n}
= 3^{-3n}left(1+frac1{3n}right)^{-3n}3^{3n}=left(1+frac1{3n}right)^{-3n}
to e^{-1}.
$$
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
answered Nov 22 '18 at 17:38
FedericoFederico
4,829514
4,829514
add a comment |
add a comment |
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Central limit theorem? Where?
– Federico
Nov 21 '18 at 19:48
Do you know anything about $left(a+frac1nright)^n?$
– Federico
Nov 21 '18 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 '18 at 19:49
@Federico Yes, but I don't see how to apply it here.
– Bogdan Vlad
Nov 21 '18 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 '18 at 22:00