Limit $ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$
up vote
3
down vote
favorite
I'm required to compute the limit of the following sequence:
$$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$
I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.
Let $a_n$ denote the base.
$$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$
From there though I don't know how to compute:
$$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$
I tried using the ln function though I couldn't get any further. I'd be glad for help :)
real-analysis calculus limits
edited Dec 5 at 13:26
Arjang
5,56662363
asked Dec 5 at 10:47
Moshe
376
add a comment |
up vote
3
down vote
favorite
I'm required to compute the limit of the following sequence:
$$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$
I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.
Let $a_n$ denote the base.
$$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$
From there though I don't know how to compute:
$$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$
I tried using the ln function though I couldn't get any further. I'd be glad for help :)
real-analysis calculus limits
edited Dec 5 at 13:26
Arjang
5,56662363
asked Dec 5 at 10:47
Moshe
376
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm required to compute the limit of the following sequence:
$$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$
I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.
Let $a_n$ denote the base.
$$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$
From there though I don't know how to compute:
$$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$
I tried using the ln function though I couldn't get any further. I'd be glad for help :)
real-analysis calculus limits
edited Dec 5 at 13:26
Arjang
5,56662363
asked Dec 5 at 10:47
Moshe
376
I'm required to compute the limit of the following sequence:
$$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$
I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.
Let $a_n$ denote the base.
$$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$
From there though I don't know how to compute:
$$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$
I tried using the ln function though I couldn't get any further. I'd be glad for help :)
real-analysis calculus limits
real-analysis calculus limits
edited Dec 5 at 13:26
Arjang
5,56662363
asked Dec 5 at 10:47
Moshe
376
edited Dec 5 at 13:26
Arjang
5,56662363
asked Dec 5 at 10:47
Moshe
376
edited Dec 5 at 13:26
Arjang
5,56662363
edited Dec 5 at 13:26
Arjang
5,56662363
edited Dec 5 at 13:26
Arjang
5,56662363
5,56662363
asked Dec 5 at 10:47
Moshe
376
asked Dec 5 at 10:47
Moshe
376
asked Dec 5 at 10:47
Moshe
376
376
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
HINT
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$
indeed
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$
and
$$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$
answered Dec 5 at 10:49
gimusi
92.9k94495
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
|
show 2 more comments
up vote
6
down vote
You can estimate:
$$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$
$$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$
answered Dec 5 at 12:15
farruhota
18.7k2736
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
Thank you gimusi.
– farruhota
Dec 5 at 13:01
add a comment |
up vote
0
down vote
You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.
answered Dec 5 at 14:14
AmbretteOrrisey
54110
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
HINT
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$
indeed
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$
and
$$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$
answered Dec 5 at 10:49
gimusi
92.9k94495
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
|
show 2 more comments
up vote
4
down vote
accepted
HINT
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$
indeed
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$
and
$$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$
answered Dec 5 at 10:49
gimusi
92.9k94495
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
|
show 2 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
HINT
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$
indeed
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$
and
$$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$
answered Dec 5 at 10:49
gimusi
92.9k94495
HINT
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$
indeed
$$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$
and
$$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$
answered Dec 5 at 10:49
gimusi
92.9k94495
edited Dec 5 at 15:13
answered Dec 5 at 10:49
gimusi
92.9k94495
answered Dec 5 at 10:49
gimusi
92.9k94495
answered Dec 5 at 10:49
gimusi
92.9k94495
92.9k94495
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
|
show 2 more comments
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
– Moshe
Dec 5 at 11:05
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
@Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
– gimusi
Dec 5 at 11:16
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
Nice and simple (as usual !). Cheers.
– Claude Leibovici
Dec 5 at 11:25
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
@ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
– gimusi
Dec 5 at 11:28
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
"[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
– Did
Dec 5 at 15:01
|
show 2 more comments
up vote
6
down vote
You can estimate:
$$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$
$$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$
answered Dec 5 at 12:15
farruhota
18.7k2736
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
Thank you gimusi.
– farruhota
Dec 5 at 13:01
add a comment |
up vote
6
down vote
You can estimate:
$$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$
$$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$
answered Dec 5 at 12:15
farruhota
18.7k2736
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
Thank you gimusi.
– farruhota
Dec 5 at 13:01
add a comment |
up vote
6
down vote
up vote
6
down vote
You can estimate:
$$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$
$$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$
answered Dec 5 at 12:15
farruhota
18.7k2736
You can estimate:
$$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$
$$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$
answered Dec 5 at 12:15
farruhota
18.7k2736
answered Dec 5 at 12:15
farruhota
18.7k2736
answered Dec 5 at 12:15
farruhota
18.7k2736
answered Dec 5 at 12:15
farruhota
18.7k2736
18.7k2736
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
Thank you gimusi.
– farruhota
Dec 5 at 13:01
add a comment |
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
Thank you gimusi.
– farruhota
Dec 5 at 13:01
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
very nice alternative approach (+1).
– gimusi
Dec 5 at 12:58
Thank you gimusi.
– farruhota
Dec 5 at 13:01
Thank you gimusi.
– farruhota
Dec 5 at 13:01
add a comment |
up vote
0
down vote
You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.
answered Dec 5 at 14:14
AmbretteOrrisey
54110
add a comment |
up vote
0
down vote
You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.
answered Dec 5 at 14:14
AmbretteOrrisey
54110
add a comment |
up vote
0
down vote
up vote
0
down vote
You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.
answered Dec 5 at 14:14
AmbretteOrrisey
54110
You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.
answered Dec 5 at 14:14
AmbretteOrrisey
54110
answered Dec 5 at 14:14
AmbretteOrrisey
54110
answered Dec 5 at 14:14
AmbretteOrrisey
54110
answered Dec 5 at 14:14
AmbretteOrrisey
54110
54110
add a comment |
add a comment |
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