Mixing
Where to start with: $lim_{ntoinfty} (sqrt[3]{n^{48}+n} - sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
$begingroup$
I have limit:
$lim_{ntoinfty} (sqrt[3]{n^{48}+n} - sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube roots by multiply them with
$dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}$
which gives me:
$dfrac{((n^{48}+n) - (n^{48}+n^2))((n^3 +3)^{12} - (n^4+4n)^9 )}{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
but I can not move from here other than just do
$dfrac{(n - n^2)((n^3 +3)^{12} - (n^4+4n)^9) }{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
I am trying to find some known limits there or something I can grasp on, but can not find anything.
real-analysis limits radicals
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 15:01
cris14cris14
1338
$endgroup$
|
show 2 more comments
$begingroup$
I have limit:
$lim_{ntoinfty} (sqrt[3]{n^{48}+n} - sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube roots by multiply them with
$dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}$
which gives me:
$dfrac{((n^{48}+n) - (n^{48}+n^2))((n^3 +3)^{12} - (n^4+4n)^9 )}{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
but I can not move from here other than just do
$dfrac{(n - n^2)((n^3 +3)^{12} - (n^4+4n)^9) }{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
I am trying to find some known limits there or something I can grasp on, but can not find anything.
real-analysis limits radicals
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 15:01
cris14cris14
1338
$endgroup$
$begingroup$
@JyrkiLahtonen But that's what I did when multiplying with $dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2})$ or not?
$endgroup$
– cris14
Jan 29 at 15:12
$begingroup$
Oh. Sorry. I misread.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:17
$begingroup$
@JyrkiLahtonen No worries, I have been wondering if it is at least a good start? Because the formula I get after the multiplying is looking like it is getting me nowhere.
$endgroup$
– cris14
Jan 29 at 15:21
$begingroup$
You still need to (at least) observe that the binomial formula expansions of both $(n^3+3)^{12}$ and $(n^4+4n)^9$ begin with $n^{36}+36n^{33}+cdots$, so those terms cancel. The next terms on the other hand...
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:22
$begingroup$
But, yes, that's a good start.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:23
|
show 2 more comments
$begingroup$
I have limit:
$lim_{ntoinfty} (sqrt[3]{n^{48}+n} - sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube roots by multiply them with
$dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}$
which gives me:
$dfrac{((n^{48}+n) - (n^{48}+n^2))((n^3 +3)^{12} - (n^4+4n)^9 )}{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
but I can not move from here other than just do
$dfrac{(n - n^2)((n^3 +3)^{12} - (n^4+4n)^9) }{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
I am trying to find some known limits there or something I can grasp on, but can not find anything.
real-analysis limits radicals
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 15:01
cris14cris14
1338
$endgroup$
I have limit:
$lim_{ntoinfty} (sqrt[3]{n^{48}+n} - sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube roots by multiply them with
$dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2)}$
which gives me:
$dfrac{((n^{48}+n) - (n^{48}+n^2))((n^3 +3)^{12} - (n^4+4n)^9 )}{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
but I can not move from here other than just do
$dfrac{(n - n^2)((n^3 +3)^{12} - (n^4+4n)^9) }{(sqrt[3]{1n^{48}+n})^2 + (sqrt[3]{1n^{48}+n})(sqrt[3]{1n^{48}+n^2})+ (sqrt[3]{1n^{48}+n^2})^2)}$
I am trying to find some known limits there or something I can grasp on, but can not find anything.
real-analysis limits radicals
real-analysis limits radicals
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 15:01
cris14cris14
1338
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
asked Jan 29 at 15:01
cris14cris14
1338
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
edited Jan 29 at 16:07
Martin Sleziak
44.9k10122277
44.9k10122277
asked Jan 29 at 15:01
cris14cris14
1338
asked Jan 29 at 15:01
cris14cris14
1338
asked Jan 29 at 15:01
cris14cris14
1338
1338
$begingroup$
@JyrkiLahtonen But that's what I did when multiplying with $dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2})$ or not?
$endgroup$
– cris14
Jan 29 at 15:12
$begingroup$
Oh. Sorry. I misread.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:17
$begingroup$
@JyrkiLahtonen No worries, I have been wondering if it is at least a good start? Because the formula I get after the multiplying is looking like it is getting me nowhere.
$endgroup$
– cris14
Jan 29 at 15:21
$begingroup$
You still need to (at least) observe that the binomial formula expansions of both $(n^3+3)^{12}$ and $(n^4+4n)^9$ begin with $n^{36}+36n^{33}+cdots$, so those terms cancel. The next terms on the other hand...
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:22
$begingroup$
But, yes, that's a good start.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:23
|
show 2 more comments
$begingroup$
@JyrkiLahtonen But that's what I did when multiplying with $dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2})$ or not?
$endgroup$
– cris14
Jan 29 at 15:12
$begingroup$
Oh. Sorry. I misread.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:17
$begingroup$
@JyrkiLahtonen No worries, I have been wondering if it is at least a good start? Because the formula I get after the multiplying is looking like it is getting me nowhere.
$endgroup$
– cris14
Jan 29 at 15:21
$begingroup$
You still need to (at least) observe that the binomial formula expansions of both $(n^3+3)^{12}$ and $(n^4+4n)^9$ begin with $n^{36}+36n^{33}+cdots$, so those terms cancel. The next terms on the other hand...
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:22
$begingroup$
But, yes, that's a good start.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:23
$begingroup$
@JyrkiLahtonen But that's what I did when multiplying with $dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2})$ or not?
$endgroup$
– cris14
Jan 29 at 15:12
$begingroup$
@JyrkiLahtonen But that's what I did when multiplying with $dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2})$ or not?
$endgroup$
– cris14
Jan 29 at 15:12
$begingroup$
Oh. Sorry. I misread.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:17
$begingroup$
Oh. Sorry. I misread.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:17
$begingroup$
@JyrkiLahtonen No worries, I have been wondering if it is at least a good start? Because the formula I get after the multiplying is looking like it is getting me nowhere.
$endgroup$
– cris14
Jan 29 at 15:21
$begingroup$
@JyrkiLahtonen No worries, I have been wondering if it is at least a good start? Because the formula I get after the multiplying is looking like it is getting me nowhere.
$endgroup$
– cris14
Jan 29 at 15:21
$begingroup$
You still need to (at least) observe that the binomial formula expansions of both $(n^3+3)^{12}$ and $(n^4+4n)^9$ begin with $n^{36}+36n^{33}+cdots$, so those terms cancel. The next terms on the other hand...
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:22
$begingroup$
You still need to (at least) observe that the binomial formula expansions of both $(n^3+3)^{12}$ and $(n^4+4n)^9$ begin with $n^{36}+36n^{33}+cdots$, so those terms cancel. The next terms on the other hand...
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:22
$begingroup$
But, yes, that's a good start.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:23
$begingroup$
But, yes, that's a good start.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:23
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\
=((n^{36}+36n^{33}+594n^{30}+cdots)-(n^{36}+36n^{33}+576n^{30}+cdots))\
=18n^{30}+cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(sqrt[3]{n^{48}+n}-sqrt[3]{n^{48}+n^{2}})\
=n^{16}(sqrt[3]{1+n^{-47}}-sqrt[3]{1+n^{-46}})\
=n^{16}((1+frac{1}{3}n^{-47}+cdots)-(1+frac{1}{3}n^{-46}+cdots))\
=n^{16}(-frac{1}{3}n^{-46}+frac{1}{3}n^{-47}+cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+cdots)$$
which equals $-6$ in the limit.
Hope that helps
answered Jan 29 at 17:19
xidgelxidgel
28815
$endgroup$
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1 Answer
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$begingroup$
Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\
=((n^{36}+36n^{33}+594n^{30}+cdots)-(n^{36}+36n^{33}+576n^{30}+cdots))\
=18n^{30}+cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(sqrt[3]{n^{48}+n}-sqrt[3]{n^{48}+n^{2}})\
=n^{16}(sqrt[3]{1+n^{-47}}-sqrt[3]{1+n^{-46}})\
=n^{16}((1+frac{1}{3}n^{-47}+cdots)-(1+frac{1}{3}n^{-46}+cdots))\
=n^{16}(-frac{1}{3}n^{-46}+frac{1}{3}n^{-47}+cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+cdots)$$
which equals $-6$ in the limit.
Hope that helps
answered Jan 29 at 17:19
xidgelxidgel
28815
$endgroup$
add a comment |
$begingroup$
Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\
=((n^{36}+36n^{33}+594n^{30}+cdots)-(n^{36}+36n^{33}+576n^{30}+cdots))\
=18n^{30}+cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(sqrt[3]{n^{48}+n}-sqrt[3]{n^{48}+n^{2}})\
=n^{16}(sqrt[3]{1+n^{-47}}-sqrt[3]{1+n^{-46}})\
=n^{16}((1+frac{1}{3}n^{-47}+cdots)-(1+frac{1}{3}n^{-46}+cdots))\
=n^{16}(-frac{1}{3}n^{-46}+frac{1}{3}n^{-47}+cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+cdots)$$
which equals $-6$ in the limit.
Hope that helps
answered Jan 29 at 17:19
xidgelxidgel
28815
$endgroup$
add a comment |
$begingroup$
Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\
=((n^{36}+36n^{33}+594n^{30}+cdots)-(n^{36}+36n^{33}+576n^{30}+cdots))\
=18n^{30}+cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(sqrt[3]{n^{48}+n}-sqrt[3]{n^{48}+n^{2}})\
=n^{16}(sqrt[3]{1+n^{-47}}-sqrt[3]{1+n^{-46}})\
=n^{16}((1+frac{1}{3}n^{-47}+cdots)-(1+frac{1}{3}n^{-46}+cdots))\
=n^{16}(-frac{1}{3}n^{-46}+frac{1}{3}n^{-47}+cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+cdots)$$
which equals $-6$ in the limit.
Hope that helps
answered Jan 29 at 17:19
xidgelxidgel
28815
$endgroup$
Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\
=((n^{36}+36n^{33}+594n^{30}+cdots)-(n^{36}+36n^{33}+576n^{30}+cdots))\
=18n^{30}+cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(sqrt[3]{n^{48}+n}-sqrt[3]{n^{48}+n^{2}})\
=n^{16}(sqrt[3]{1+n^{-47}}-sqrt[3]{1+n^{-46}})\
=n^{16}((1+frac{1}{3}n^{-47}+cdots)-(1+frac{1}{3}n^{-46}+cdots))\
=n^{16}(-frac{1}{3}n^{-46}+frac{1}{3}n^{-47}+cdots)$$
The product of the two terms is:
$$(-6+6n^{-1}+cdots)$$
which equals $-6$ in the limit.
Hope that helps
answered Jan 29 at 17:19
xidgelxidgel
28815
answered Jan 29 at 17:19
xidgelxidgel
28815
answered Jan 29 at 17:19
xidgelxidgel
28815
answered Jan 29 at 17:19
xidgelxidgel
28815
28815
add a comment |
add a comment |
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$begingroup$
@JyrkiLahtonen But that's what I did when multiplying with $dfrac{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2}{(sqrt[3]{n^{48}+n})^2 + (sqrt[3]{n^{48}+n})(sqrt[3]{n^{48}+n^2})+ (sqrt[3]{n^{48}+n^2})^2})$ or not?
$endgroup$
– cris14
Jan 29 at 15:12
$begingroup$
Oh. Sorry. I misread.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:17
$begingroup$
@JyrkiLahtonen No worries, I have been wondering if it is at least a good start? Because the formula I get after the multiplying is looking like it is getting me nowhere.
$endgroup$
– cris14
Jan 29 at 15:21
$begingroup$
You still need to (at least) observe that the binomial formula expansions of both $(n^3+3)^{12}$ and $(n^4+4n)^9$ begin with $n^{36}+36n^{33}+cdots$, so those terms cancel. The next terms on the other hand...
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:22
$begingroup$
But, yes, that's a good start.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 15:23