Find the limit of $lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity...
Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
sequences-and-series
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
asked Dec 22 at 9:41
Belkan
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
sequences-and-series
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
asked Dec 22 at 9:41
Belkan
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That's a sequence, not a series.
– José Carlos Santos
Dec 22 at 9:45
Oops, corrected it.
– Belkan
Dec 22 at 9:47
add a comment |
Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
sequences-and-series
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
asked Dec 22 at 9:41
Belkan
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
sequences-and-series
sequences-and-series
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
asked Dec 22 at 9:41
Belkan
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
asked Dec 22 at 9:41
Belkan
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
edited Dec 22 at 11:22
Asaf Karagila♦
301k32423755
301k32423755
asked Dec 22 at 9:41
Belkan
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 22 at 9:41
Belkan
587
asked Dec 22 at 9:41
Belkan
587
587
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Belkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That's a sequence, not a series.
– José Carlos Santos
Dec 22 at 9:45
Oops, corrected it.
– Belkan
Dec 22 at 9:47
add a comment |
That's a sequence, not a series.
– José Carlos Santos
Dec 22 at 9:45
Oops, corrected it.
– Belkan
Dec 22 at 9:47
That's a sequence, not a series.
– José Carlos Santos
Dec 22 at 9:45
That's a sequence, not a series.
– José Carlos Santos
Dec 22 at 9:45
Oops, corrected it.
– Belkan
Dec 22 at 9:47
Oops, corrected it.
– Belkan
Dec 22 at 9:47
add a comment |
3 Answers
3
active
oldest
votes
$$
begin{align}
lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
&=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
&=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
&=frac13
end{align}
$$
answered Dec 22 at 9:50
robjohn♦
264k27303623
add a comment |
My favorite way is to consider
$$
f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
=frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
$$
g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
answered Dec 22 at 10:01
egreg
177k1484200
add a comment |
1) $n(1+1/n )^{1/3} = $
$dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$small{lim_{n rightarrow infty}(1/n^2)×}$
$small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$small{0 cdot (1/3)=0.}$
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
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
$$
begin{align}
lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
&=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
&=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
&=frac13
end{align}
$$
answered Dec 22 at 9:50
robjohn♦
264k27303623
add a comment |
$$
begin{align}
lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
&=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
&=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
&=frac13
end{align}
$$
answered Dec 22 at 9:50
robjohn♦
264k27303623
add a comment |
$$
begin{align}
lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
&=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
&=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
&=frac13
end{align}
$$
answered Dec 22 at 9:50
robjohn♦
264k27303623
$$
begin{align}
lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
&=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
&=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
&=frac13
end{align}
$$
answered Dec 22 at 9:50
robjohn♦
264k27303623
answered Dec 22 at 9:50
robjohn♦
264k27303623
answered Dec 22 at 9:50
robjohn♦
264k27303623
answered Dec 22 at 9:50
robjohn♦
264k27303623
264k27303623
add a comment |
add a comment |
My favorite way is to consider
$$
f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
=frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
$$
g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
answered Dec 22 at 10:01
egreg
177k1484200
add a comment |
My favorite way is to consider
$$
f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
=frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
$$
g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
answered Dec 22 at 10:01
egreg
177k1484200
add a comment |
My favorite way is to consider
$$
f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
=frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
$$
g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
answered Dec 22 at 10:01
egreg
177k1484200
My favorite way is to consider
$$
f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
=frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
$$
g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
answered Dec 22 at 10:01
egreg
177k1484200
answered Dec 22 at 10:01
egreg
177k1484200
answered Dec 22 at 10:01
egreg
177k1484200
answered Dec 22 at 10:01
egreg
177k1484200
177k1484200
add a comment |
add a comment |
1) $n(1+1/n )^{1/3} = $
$dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$small{lim_{n rightarrow infty}(1/n^2)×}$
$small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$small{0 cdot (1/3)=0.}$
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
add a comment |
1) $n(1+1/n )^{1/3} = $
$dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$small{lim_{n rightarrow infty}(1/n^2)×}$
$small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$small{0 cdot (1/3)=0.}$
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
add a comment |
1) $n(1+1/n )^{1/3} = $
$dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$small{lim_{n rightarrow infty}(1/n^2)×}$
$small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$small{0 cdot (1/3)=0.}$
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
1) $n(1+1/n )^{1/3} = $
$dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$small{lim_{n rightarrow infty}(1/n^2)×}$
$small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$small{0 cdot (1/3)=0.}$
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
edited Dec 22 at 10:56
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
answered Dec 22 at 10:30
Peter Szilas
10.6k2720
10.6k2720
add a comment |
add a comment |
Belkan is a new contributor. Be nice, and check out our Code of Conduct.
Belkan is a new contributor. Be nice, and check out our Code of Conduct.
Belkan is a new contributor. Be nice, and check out our Code of Conduct.
Belkan is a new contributor. Be nice, and check out our Code of Conduct.
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That's a sequence, not a series.
– José Carlos Santos
Dec 22 at 9:45
Oops, corrected it.
– Belkan
Dec 22 at 9:47