Simple limit of a sequence with triple logarithm
$begingroup$
This is not a homework! :) Just preparing for exam.
How can I calculate the following limit of a sequence?
$$ lim_{n to infty } left[4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1)right]$$
Should I first take into account only the first two components (dividing logarithms) and then take into account the final component (divide logarithms again)?
The kind of funny thing is that I can easily solve sort of more complex limits and simpler limits, I'm specifically struggling with this one (and other logarithmic limits).
Thanks.
real-analysis limits
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
asked Dec 29 '18 at 3:09
wenoweno
29211
$endgroup$
add a comment |
$begingroup$
This is not a homework! :) Just preparing for exam.
How can I calculate the following limit of a sequence?
$$ lim_{n to infty } left[4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1)right]$$
Should I first take into account only the first two components (dividing logarithms) and then take into account the final component (divide logarithms again)?
The kind of funny thing is that I can easily solve sort of more complex limits and simpler limits, I'm specifically struggling with this one (and other logarithmic limits).
Thanks.
real-analysis limits
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
asked Dec 29 '18 at 3:09
wenoweno
29211
$endgroup$
add a comment |
$begingroup$
This is not a homework! :) Just preparing for exam.
How can I calculate the following limit of a sequence?
$$ lim_{n to infty } left[4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1)right]$$
Should I first take into account only the first two components (dividing logarithms) and then take into account the final component (divide logarithms again)?
The kind of funny thing is that I can easily solve sort of more complex limits and simpler limits, I'm specifically struggling with this one (and other logarithmic limits).
Thanks.
real-analysis limits
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
asked Dec 29 '18 at 3:09
wenoweno
29211
$endgroup$
This is not a homework! :) Just preparing for exam.
How can I calculate the following limit of a sequence?
$$ lim_{n to infty } left[4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1)right]$$
Should I first take into account only the first two components (dividing logarithms) and then take into account the final component (divide logarithms again)?
The kind of funny thing is that I can easily solve sort of more complex limits and simpler limits, I'm specifically struggling with this one (and other logarithmic limits).
Thanks.
real-analysis limits
real-analysis limits
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
asked Dec 29 '18 at 3:09
wenoweno
29211
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
asked Dec 29 '18 at 3:09
wenoweno
29211
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
edited Dec 29 '18 at 3:10
Chinnapparaj R
5,5422928
5,5422928
asked Dec 29 '18 at 3:09
wenoweno
29211
asked Dec 29 '18 at 3:09
wenoweno
29211
asked Dec 29 '18 at 3:09
wenoweno
29211
29211
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = log(16n^{4}) - log(2n^2-1) - log(3n^2-1) +log(3)}$$
and so
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = logleft(frac{16n^4}{(2n^2-1)(3n^2-1)}right)+log(3)}$$
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
add a comment |
$begingroup$
As usual, there are many ways to do it.
Remember that $n$ is large; so
$$4log(2{n})-logleft(frac{2{n}^2 - 1}{3}right) - log(3n^2-1)simeq 4log(2{n})-logleft(frac{2{n}^2}{3}right) - log(3n^2)$$ Now, either combine the logarithms as Lucas Corrêa did or just expand them as
$$4 log(2)+4log(n)-log(2)-2log(n)+log(3)-log(3)-2log(n)$$ and simplify.
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = log(16n^{4}) - log(2n^2-1) - log(3n^2-1) +log(3)}$$
and so
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = logleft(frac{16n^4}{(2n^2-1)(3n^2-1)}right)+log(3)}$$
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
add a comment |
$begingroup$
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = log(16n^{4}) - log(2n^2-1) - log(3n^2-1) +log(3)}$$
and so
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = logleft(frac{16n^4}{(2n^2-1)(3n^2-1)}right)+log(3)}$$
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
add a comment |
$begingroup$
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = log(16n^{4}) - log(2n^2-1) - log(3n^2-1) +log(3)}$$
and so
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = logleft(frac{16n^4}{(2n^2-1)(3n^2-1)}right)+log(3)}$$
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = log(16n^{4}) - log(2n^2-1) - log(3n^2-1) +log(3)}$$
and so
$${small 4log(2text{n})-logleft(frac{2text{n}^2 - 1}{3}right) - log(3n^2-1) = logleft(frac{16n^4}{(2n^2-1)(3n^2-1)}right)+log(3)}$$
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
edited Dec 29 '18 at 7:36
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
answered Dec 29 '18 at 3:23
Lucas CorrêaLucas Corrêa
1,5751321
1,5751321
add a comment |
add a comment |
$begingroup$
As usual, there are many ways to do it.
Remember that $n$ is large; so
$$4log(2{n})-logleft(frac{2{n}^2 - 1}{3}right) - log(3n^2-1)simeq 4log(2{n})-logleft(frac{2{n}^2}{3}right) - log(3n^2)$$ Now, either combine the logarithms as Lucas Corrêa did or just expand them as
$$4 log(2)+4log(n)-log(2)-2log(n)+log(3)-log(3)-2log(n)$$ and simplify.
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
As usual, there are many ways to do it.
Remember that $n$ is large; so
$$4log(2{n})-logleft(frac{2{n}^2 - 1}{3}right) - log(3n^2-1)simeq 4log(2{n})-logleft(frac{2{n}^2}{3}right) - log(3n^2)$$ Now, either combine the logarithms as Lucas Corrêa did or just expand them as
$$4 log(2)+4log(n)-log(2)-2log(n)+log(3)-log(3)-2log(n)$$ and simplify.
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
As usual, there are many ways to do it.
Remember that $n$ is large; so
$$4log(2{n})-logleft(frac{2{n}^2 - 1}{3}right) - log(3n^2-1)simeq 4log(2{n})-logleft(frac{2{n}^2}{3}right) - log(3n^2)$$ Now, either combine the logarithms as Lucas Corrêa did or just expand them as
$$4 log(2)+4log(n)-log(2)-2log(n)+log(3)-log(3)-2log(n)$$ and simplify.
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
As usual, there are many ways to do it.
Remember that $n$ is large; so
$$4log(2{n})-logleft(frac{2{n}^2 - 1}{3}right) - log(3n^2-1)simeq 4log(2{n})-logleft(frac{2{n}^2}{3}right) - log(3n^2)$$ Now, either combine the logarithms as Lucas Corrêa did or just expand them as
$$4 log(2)+4log(n)-log(2)-2log(n)+log(3)-log(3)-2log(n)$$ and simplify.
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
answered Dec 29 '18 at 4:01
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
add a comment |
add a comment |
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