How to resolve $lim_xtoinfty (1+frac1x^n)^x$
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So I was interested in limits of the following form
$$lim_xtoinfty (1+frac1x^n)^x qquad ngeqslant1$$
when $n=1$, the limit is $e$, of course. I was then able to prove the limit approaches $1$ when $ngeqslant2$. I cannot figure out how deal with the case $1lt n lt 2$. I know it converges, but I can't figure out where it converges.
calculus limits
asked Aug 16 at 4:53
Mark Abadir
11
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up vote
0
down vote
favorite
So I was interested in limits of the following form
$$lim_xtoinfty (1+frac1x^n)^x qquad ngeqslant1$$
when $n=1$, the limit is $e$, of course. I was then able to prove the limit approaches $1$ when $ngeqslant2$. I cannot figure out how deal with the case $1lt n lt 2$. I know it converges, but I can't figure out where it converges.
calculus limits
asked Aug 16 at 4:53
Mark Abadir
11
$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1$.
– Riemann
Aug 16 at 4:57
Can you adapt your method for $n=2$ to work for any $n>1$?
– Lord Shark the Unknown
Aug 16 at 4:57
@Riemann I don't quite see where to do from there
– Mark Abadir
Aug 16 at 5:03
If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
– Riemann
Aug 16 at 5:06
@LordSharktheUnknown my method for $n=2$ was factoring what was inside the limit and resolving the factors as $exp(ix)exp(-ix) = 1$. From there I used limit comparison since I knew $frac1x^n leq frac1x^2$ for $ngeq 2$. It doesn't apply for $nlt 2$
– Mark Abadir
Aug 16 at 5:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I was interested in limits of the following form
$$lim_xtoinfty (1+frac1x^n)^x qquad ngeqslant1$$
when $n=1$, the limit is $e$, of course. I was then able to prove the limit approaches $1$ when $ngeqslant2$. I cannot figure out how deal with the case $1lt n lt 2$. I know it converges, but I can't figure out where it converges.
calculus limits
asked Aug 16 at 4:53
Mark Abadir
11
So I was interested in limits of the following form
$$lim_xtoinfty (1+frac1x^n)^x qquad ngeqslant1$$
when $n=1$, the limit is $e$, of course. I was then able to prove the limit approaches $1$ when $ngeqslant2$. I cannot figure out how deal with the case $1lt n lt 2$. I know it converges, but I can't figure out where it converges.
calculus limits
asked Aug 16 at 4:53
Mark Abadir
11
asked Aug 16 at 4:53
Mark Abadir
11
asked Aug 16 at 4:53
Mark Abadir
11
asked Aug 16 at 4:53
Mark Abadir
11
11
$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1$.
– Riemann
Aug 16 at 4:57
Can you adapt your method for $n=2$ to work for any $n>1$?
– Lord Shark the Unknown
Aug 16 at 4:57
@Riemann I don't quite see where to do from there
– Mark Abadir
Aug 16 at 5:03
If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
– Riemann
Aug 16 at 5:06
@LordSharktheUnknown my method for $n=2$ was factoring what was inside the limit and resolving the factors as $exp(ix)exp(-ix) = 1$. From there I used limit comparison since I knew $frac1x^n leq frac1x^2$ for $ngeq 2$. It doesn't apply for $nlt 2$
– Mark Abadir
Aug 16 at 5:07
add a comment |Â
$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1$.
– Riemann
Aug 16 at 4:57
Can you adapt your method for $n=2$ to work for any $n>1$?
– Lord Shark the Unknown
Aug 16 at 4:57
@Riemann I don't quite see where to do from there
– Mark Abadir
Aug 16 at 5:03
If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
– Riemann
Aug 16 at 5:06
@LordSharktheUnknown my method for $n=2$ was factoring what was inside the limit and resolving the factors as $exp(ix)exp(-ix) = 1$. From there I used limit comparison since I knew $frac1x^n leq frac1x^2$ for $ngeq 2$. It doesn't apply for $nlt 2$
– Mark Abadir
Aug 16 at 5:07
$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1$.
– Riemann
Aug 16 at 4:57
$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1$.
– Riemann
Aug 16 at 4:57
Can you adapt your method for $n=2$ to work for any $n>1$?
– Lord Shark the Unknown
Aug 16 at 4:57
Can you adapt your method for $n=2$ to work for any $n>1$?
– Lord Shark the Unknown
Aug 16 at 4:57
@Riemann I don't quite see where to do from there
– Mark Abadir
Aug 16 at 5:03
@Riemann I don't quite see where to do from there
– Mark Abadir
Aug 16 at 5:03
If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
– Riemann
Aug 16 at 5:06
If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
– Riemann
Aug 16 at 5:06
@LordSharktheUnknown my method for $n=2$ was factoring what was inside the limit and resolving the factors as $exp(ix)exp(-ix) = 1$. From there I used limit comparison since I knew $frac1x^n leq frac1x^2$ for $ngeq 2$. It doesn't apply for $nlt 2$
– Mark Abadir
Aug 16 at 5:07
@LordSharktheUnknown my method for $n=2$ was factoring what was inside the limit and resolving the factors as $exp(ix)exp(-ix) = 1$. From there I used limit comparison since I knew $frac1x^n leq frac1x^2$ for $ngeq 2$. It doesn't apply for $nlt 2$
– Mark Abadir
Aug 16 at 5:07
add a comment |Â
2 Answers
2
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up vote
2
down vote
Consider $$a= left(1+frac1x^nright)^ximplies log(a)=x log left(1+frac1x^nright)$$ Since $x$ is large, using equivalents
$$log(a) sim frac1x^n-1$$ which goes to $0$ if $n>1$. So $a to 1$.
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
add a comment |Â
up vote
1
down vote
When $n=1$ it is easy to do!
Because $$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1,n>1.$$
And use the result:If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
In this case, let $a_n=(1+frac1x^n)^x^n$ and $b_n=frac1x^n-1$, so
$a_nto e$ and $b_nto 0$, and the desired limit is $1$.
answered Aug 16 at 5:12
Riemann
2,4811219
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Consider $$a= left(1+frac1x^nright)^ximplies log(a)=x log left(1+frac1x^nright)$$ Since $x$ is large, using equivalents
$$log(a) sim frac1x^n-1$$ which goes to $0$ if $n>1$. So $a to 1$.
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
add a comment |Â
up vote
2
down vote
Consider $$a= left(1+frac1x^nright)^ximplies log(a)=x log left(1+frac1x^nright)$$ Since $x$ is large, using equivalents
$$log(a) sim frac1x^n-1$$ which goes to $0$ if $n>1$. So $a to 1$.
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Consider $$a= left(1+frac1x^nright)^ximplies log(a)=x log left(1+frac1x^nright)$$ Since $x$ is large, using equivalents
$$log(a) sim frac1x^n-1$$ which goes to $0$ if $n>1$. So $a to 1$.
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
Consider $$a= left(1+frac1x^nright)^ximplies log(a)=x log left(1+frac1x^nright)$$ Since $x$ is large, using equivalents
$$log(a) sim frac1x^n-1$$ which goes to $0$ if $n>1$. So $a to 1$.
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
answered Aug 16 at 5:24
Claude Leibovici
112k1055127
112k1055127
add a comment |Â
add a comment |Â
up vote
1
down vote
When $n=1$ it is easy to do!
Because $$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1,n>1.$$
And use the result:If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
In this case, let $a_n=(1+frac1x^n)^x^n$ and $b_n=frac1x^n-1$, so
$a_nto e$ and $b_nto 0$, and the desired limit is $1$.
answered Aug 16 at 5:12
Riemann
2,4811219
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
add a comment |Â
up vote
1
down vote
When $n=1$ it is easy to do!
Because $$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1,n>1.$$
And use the result:If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
In this case, let $a_n=(1+frac1x^n)^x^n$ and $b_n=frac1x^n-1$, so
$a_nto e$ and $b_nto 0$, and the desired limit is $1$.
answered Aug 16 at 5:12
Riemann
2,4811219
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
add a comment |Â
up vote
1
down vote
up vote
1
down vote
When $n=1$ it is easy to do!
Because $$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1,n>1.$$
And use the result:If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
In this case, let $a_n=(1+frac1x^n)^x^n$ and $b_n=frac1x^n-1$, so
$a_nto e$ and $b_nto 0$, and the desired limit is $1$.
answered Aug 16 at 5:12
Riemann
2,4811219
When $n=1$ it is easy to do!
Because $$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1,n>1.$$
And use the result:If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
In this case, let $a_n=(1+frac1x^n)^x^n$ and $b_n=frac1x^n-1$, so
$a_nto e$ and $b_nto 0$, and the desired limit is $1$.
answered Aug 16 at 5:12
Riemann
2,4811219
answered Aug 16 at 5:12
Riemann
2,4811219
answered Aug 16 at 5:12
Riemann
2,4811219
answered Aug 16 at 5:12
Riemann
2,4811219
2,4811219
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
add a comment |Â
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
Ahh thank you. This makes sense.
– Mark Abadir
Aug 16 at 5:16
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
If you think this answer is helpful, please give a vote. Thanks a lot.
– Riemann
Aug 16 at 5:24
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
Also accept the answer if it solves your problem
– Keen-ameteur
Aug 16 at 5:54
add a comment |Â
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$(1+frac1x^n)^x =left((1+frac1x^n)^x^nright)^frac1x^n-1$.
– Riemann
Aug 16 at 4:57
Can you adapt your method for $n=2$ to work for any $n>1$?
– Lord Shark the Unknown
Aug 16 at 4:57
@Riemann I don't quite see where to do from there
– Mark Abadir
Aug 16 at 5:03
If $a_nto a>0$ and $b_nto b$, then $a_n^b_nto a^b.$
– Riemann
Aug 16 at 5:06
@LordSharktheUnknown my method for $n=2$ was factoring what was inside the limit and resolving the factors as $exp(ix)exp(-ix) = 1$. From there I used limit comparison since I knew $frac1x^n leq frac1x^2$ for $ngeq 2$. It doesn't apply for $nlt 2$
– Mark Abadir
Aug 16 at 5:07