Finding limit from Squeeze theorem.
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I recently came across a problem which stated to find the limit of an equation through the Squeeze theorem,
$$lim_ntoinfty left(frac2n-53n+1right) ^n$$
My approach: I did the question with L'Hospital's Rule just for the sake of finding the limit,
$$log (L) = n(log(2n-5) - log(3n+1))$$
$$ log(L) = fraclog(2n-5) - log(3n+1)frac1n$$
By differentiating,
$$ log(L) = fracfrac22n-5-frac33n+1frac-1n^2$$
$$ log(L) = -frac1712$$
$$ L = e^-frac1712$$
This was the limit obtained by me. But I wasn't able to approach through Squeeze Theorem.
real-analysis sequences-and-series limits
edited Aug 26 at 12:23
Davide Morgante
2,513623
asked Aug 26 at 12:20
Sahil Silare
648
 |Â
show 1 more comment
up vote
0
down vote
favorite
I recently came across a problem which stated to find the limit of an equation through the Squeeze theorem,
$$lim_ntoinfty left(frac2n-53n+1right) ^n$$
My approach: I did the question with L'Hospital's Rule just for the sake of finding the limit,
$$log (L) = n(log(2n-5) - log(3n+1))$$
$$ log(L) = fraclog(2n-5) - log(3n+1)frac1n$$
By differentiating,
$$ log(L) = fracfrac22n-5-frac33n+1frac-1n^2$$
$$ log(L) = -frac1712$$
$$ L = e^-frac1712$$
This was the limit obtained by me. But I wasn't able to approach through Squeeze Theorem.
real-analysis sequences-and-series limits
edited Aug 26 at 12:23
Davide Morgante
2,513623
asked Aug 26 at 12:20
Sahil Silare
648
3
Are you sure of it?
– Dr. Sonnhard Graubner
Aug 26 at 12:23
2
Is'nt your general term $<(2/3)^n$?
– Lord Shark the Unknown
Aug 26 at 12:24
$log(L) = fracfrac22n-5-frac33n+1frac-1n^2 implies log(L) to - infty$
– ab123
Aug 26 at 12:26
@ab123 Oh so we can't apply L'Hospital's rule twice?
– Sahil Silare
Aug 26 at 12:28
1
The limit is $0$.
– Anastassis Kapetanakis
Aug 26 at 12:28
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I recently came across a problem which stated to find the limit of an equation through the Squeeze theorem,
$$lim_ntoinfty left(frac2n-53n+1right) ^n$$
My approach: I did the question with L'Hospital's Rule just for the sake of finding the limit,
$$log (L) = n(log(2n-5) - log(3n+1))$$
$$ log(L) = fraclog(2n-5) - log(3n+1)frac1n$$
By differentiating,
$$ log(L) = fracfrac22n-5-frac33n+1frac-1n^2$$
$$ log(L) = -frac1712$$
$$ L = e^-frac1712$$
This was the limit obtained by me. But I wasn't able to approach through Squeeze Theorem.
real-analysis sequences-and-series limits
edited Aug 26 at 12:23
Davide Morgante
2,513623
asked Aug 26 at 12:20
Sahil Silare
648
I recently came across a problem which stated to find the limit of an equation through the Squeeze theorem,
$$lim_ntoinfty left(frac2n-53n+1right) ^n$$
My approach: I did the question with L'Hospital's Rule just for the sake of finding the limit,
$$log (L) = n(log(2n-5) - log(3n+1))$$
$$ log(L) = fraclog(2n-5) - log(3n+1)frac1n$$
By differentiating,
$$ log(L) = fracfrac22n-5-frac33n+1frac-1n^2$$
$$ log(L) = -frac1712$$
$$ L = e^-frac1712$$
This was the limit obtained by me. But I wasn't able to approach through Squeeze Theorem.
real-analysis sequences-and-series limits
edited Aug 26 at 12:23
Davide Morgante
2,513623
asked Aug 26 at 12:20
Sahil Silare
648
edited Aug 26 at 12:23
Davide Morgante
2,513623
edited Aug 26 at 12:23
Davide Morgante
2,513623
edited Aug 26 at 12:23
Davide Morgante
2,513623
2,513623
asked Aug 26 at 12:20
Sahil Silare
648
asked Aug 26 at 12:20
Sahil Silare
648
asked Aug 26 at 12:20
Sahil Silare
648
648
3
Are you sure of it?
– Dr. Sonnhard Graubner
Aug 26 at 12:23
2
Is'nt your general term $<(2/3)^n$?
– Lord Shark the Unknown
Aug 26 at 12:24
$log(L) = fracfrac22n-5-frac33n+1frac-1n^2 implies log(L) to - infty$
– ab123
Aug 26 at 12:26
@ab123 Oh so we can't apply L'Hospital's rule twice?
– Sahil Silare
Aug 26 at 12:28
1
The limit is $0$.
– Anastassis Kapetanakis
Aug 26 at 12:28
 |Â
show 1 more comment
3
Are you sure of it?
– Dr. Sonnhard Graubner
Aug 26 at 12:23
2
Is'nt your general term $<(2/3)^n$?
– Lord Shark the Unknown
Aug 26 at 12:24
$log(L) = fracfrac22n-5-frac33n+1frac-1n^2 implies log(L) to - infty$
– ab123
Aug 26 at 12:26
@ab123 Oh so we can't apply L'Hospital's rule twice?
– Sahil Silare
Aug 26 at 12:28
1
The limit is $0$.
– Anastassis Kapetanakis
Aug 26 at 12:28
3
3
Are you sure of it?
– Dr. Sonnhard Graubner
Aug 26 at 12:23
Are you sure of it?
– Dr. Sonnhard Graubner
Aug 26 at 12:23
2
2
Is'nt your general term $<(2/3)^n$?
– Lord Shark the Unknown
Aug 26 at 12:24
Is'nt your general term $<(2/3)^n$?
– Lord Shark the Unknown
Aug 26 at 12:24
$log(L) = fracfrac22n-5-frac33n+1frac-1n^2 implies log(L) to - infty$
– ab123
Aug 26 at 12:26
$log(L) = fracfrac22n-5-frac33n+1frac-1n^2 implies log(L) to - infty$
– ab123
Aug 26 at 12:26
@ab123 Oh so we can't apply L'Hospital's rule twice?
– Sahil Silare
Aug 26 at 12:28
@ab123 Oh so we can't apply L'Hospital's rule twice?
– Sahil Silare
Aug 26 at 12:28
1
1
The limit is $0$.
– Anastassis Kapetanakis
Aug 26 at 12:28
The limit is $0$.
– Anastassis Kapetanakis
Aug 26 at 12:28
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
$$0<left(frac2n-53n+1right) ^n<left(frac23right)^n,$$
so
$$lim_ntoinfty left(frac2n-53n+1right) ^n=0.$$
answered Aug 26 at 12:27
Riemann
2,6951219
1
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
add a comment |Â
up vote
2
down vote
We have that
$$0leleft(frac2n-53n+1right) ^nleleft(frac2n-5+53n+1-1right) ^n=left(frac23right) ^nto 0$$
answered Aug 26 at 12:29
gimusi
70.1k73786
add a comment |Â
up vote
0
down vote
I think you've tried to use L'Hopital's rule return only the denominator, not also the numerator, has $ntoinfty$ limit $0$. That's spurious. The correct analysis is $ln Lto -infty,,Lto 0$ because of the asymptotic $(2/3)^n$ behaviour.
answered Aug 26 at 12:30
J.G.
14k11525
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
$$0<left(frac2n-53n+1right) ^n<left(frac23right)^n,$$
so
$$lim_ntoinfty left(frac2n-53n+1right) ^n=0.$$
answered Aug 26 at 12:27
Riemann
2,6951219
1
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
add a comment |Â
up vote
6
down vote
accepted
$$0<left(frac2n-53n+1right) ^n<left(frac23right)^n,$$
so
$$lim_ntoinfty left(frac2n-53n+1right) ^n=0.$$
answered Aug 26 at 12:27
Riemann
2,6951219
1
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$$0<left(frac2n-53n+1right) ^n<left(frac23right)^n,$$
so
$$lim_ntoinfty left(frac2n-53n+1right) ^n=0.$$
answered Aug 26 at 12:27
Riemann
2,6951219
$$0<left(frac2n-53n+1right) ^n<left(frac23right)^n,$$
so
$$lim_ntoinfty left(frac2n-53n+1right) ^n=0.$$
answered Aug 26 at 12:27
Riemann
2,6951219
answered Aug 26 at 12:27
Riemann
2,6951219
answered Aug 26 at 12:27
Riemann
2,6951219
answered Aug 26 at 12:27
Riemann
2,6951219
2,6951219
1
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
add a comment |Â
1
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
1
1
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
In determining that the limit was "$-frac1712$" you appear to have taken the limit as n goes to 0, not infinity.
– user247327
Aug 26 at 12:31
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
Sir can't the lower bound be $left(frac-34right)^n$ ?
– Sahil Silare
Aug 26 at 13:44
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
no, when $n$ is even, $left(frac-34right)^n>left(frac23right)^n$.
– Riemann
Aug 26 at 13:52
add a comment |Â
up vote
2
down vote
We have that
$$0leleft(frac2n-53n+1right) ^nleleft(frac2n-5+53n+1-1right) ^n=left(frac23right) ^nto 0$$
answered Aug 26 at 12:29
gimusi
70.1k73786
add a comment |Â
up vote
2
down vote
We have that
$$0leleft(frac2n-53n+1right) ^nleleft(frac2n-5+53n+1-1right) ^n=left(frac23right) ^nto 0$$
answered Aug 26 at 12:29
gimusi
70.1k73786
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have that
$$0leleft(frac2n-53n+1right) ^nleleft(frac2n-5+53n+1-1right) ^n=left(frac23right) ^nto 0$$
answered Aug 26 at 12:29
gimusi
70.1k73786
We have that
$$0leleft(frac2n-53n+1right) ^nleleft(frac2n-5+53n+1-1right) ^n=left(frac23right) ^nto 0$$
answered Aug 26 at 12:29
gimusi
70.1k73786
answered Aug 26 at 12:29
gimusi
70.1k73786
answered Aug 26 at 12:29
gimusi
70.1k73786
answered Aug 26 at 12:29
gimusi
70.1k73786
70.1k73786
add a comment |Â
add a comment |Â
up vote
0
down vote
I think you've tried to use L'Hopital's rule return only the denominator, not also the numerator, has $ntoinfty$ limit $0$. That's spurious. The correct analysis is $ln Lto -infty,,Lto 0$ because of the asymptotic $(2/3)^n$ behaviour.
answered Aug 26 at 12:30
J.G.
14k11525
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
add a comment |Â
up vote
0
down vote
I think you've tried to use L'Hopital's rule return only the denominator, not also the numerator, has $ntoinfty$ limit $0$. That's spurious. The correct analysis is $ln Lto -infty,,Lto 0$ because of the asymptotic $(2/3)^n$ behaviour.
answered Aug 26 at 12:30
J.G.
14k11525
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you've tried to use L'Hopital's rule return only the denominator, not also the numerator, has $ntoinfty$ limit $0$. That's spurious. The correct analysis is $ln Lto -infty,,Lto 0$ because of the asymptotic $(2/3)^n$ behaviour.
answered Aug 26 at 12:30
J.G.
14k11525
I think you've tried to use L'Hopital's rule return only the denominator, not also the numerator, has $ntoinfty$ limit $0$. That's spurious. The correct analysis is $ln Lto -infty,,Lto 0$ because of the asymptotic $(2/3)^n$ behaviour.
answered Aug 26 at 12:30
J.G.
14k11525
answered Aug 26 at 12:30
J.G.
14k11525
answered Aug 26 at 12:30
J.G.
14k11525
answered Aug 26 at 12:30
J.G.
14k11525
14k11525
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
add a comment |Â
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
Can you clarify what's wrong with my L'Hospital rule?
– Sahil Silare
Aug 26 at 12:31
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
@SahilSilare You can't use it with $1/0$. It's as if you tried to get the limit of $1/(1/n)$ by differentiating.
– J.G.
Aug 26 at 12:58
add a comment |Â
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3
Are you sure of it?
– Dr. Sonnhard Graubner
Aug 26 at 12:23
2
Is'nt your general term $<(2/3)^n$?
– Lord Shark the Unknown
Aug 26 at 12:24
$log(L) = fracfrac22n-5-frac33n+1frac-1n^2 implies log(L) to - infty$
– ab123
Aug 26 at 12:26
@ab123 Oh so we can't apply L'Hospital's rule twice?
– Sahil Silare
Aug 26 at 12:28
1
The limit is $0$.
– Anastassis Kapetanakis
Aug 26 at 12:28