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What is $limlimits_n rightarrow infty fracsqrtn+1 + 3sqrtn+2 - 4$?
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$$beginequation*
lim_n rightarrow infty
fracsqrtn+1 + 3sqrtn+2 - 4
endequation*$$
I know that I should multiply by the conjugate if I have square roots in either the numerator or the denominator but what if I have square roots in both numerator and denominator?
Could anyone help me please?
calculus real-analysis sequences-and-series
edited Sep 2 at 3:39
Robson
47320
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$$beginequation*
lim_n rightarrow infty
fracsqrtn+1 + 3sqrtn+2 - 4
endequation*$$
I know that I should multiply by the conjugate if I have square roots in either the numerator or the denominator but what if I have square roots in both numerator and denominator?
Could anyone help me please?
calculus real-analysis sequences-and-series
edited Sep 2 at 3:39
Robson
47320
1
Both the numerator and the denominator are $sqrtn+O(1)$, so...
– Jack D'Aurizio♦
Sep 1 at 23:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$beginequation*
lim_n rightarrow infty
fracsqrtn+1 + 3sqrtn+2 - 4
endequation*$$
I know that I should multiply by the conjugate if I have square roots in either the numerator or the denominator but what if I have square roots in both numerator and denominator?
Could anyone help me please?
calculus real-analysis sequences-and-series
edited Sep 2 at 3:39
Robson
47320
$$beginequation*
lim_n rightarrow infty
fracsqrtn+1 + 3sqrtn+2 - 4
endequation*$$
I know that I should multiply by the conjugate if I have square roots in either the numerator or the denominator but what if I have square roots in both numerator and denominator?
Could anyone help me please?
calculus real-analysis sequences-and-series
edited Sep 2 at 3:39
Robson
47320
edited Sep 2 at 3:39
Robson
47320
edited Sep 2 at 3:39
Robson
47320
edited Sep 2 at 3:39
Robson
47320
47320
asked Sep 1 at 22:26
user426277
1
Both the numerator and the denominator are $sqrtn+O(1)$, so...
– Jack D'Aurizio♦
Sep 1 at 23:33
add a comment |Â
1
Both the numerator and the denominator are $sqrtn+O(1)$, so...
– Jack D'Aurizio♦
Sep 1 at 23:33
1
1
Both the numerator and the denominator are $sqrtn+O(1)$, so...
– Jack D'Aurizio♦
Sep 1 at 23:33
Both the numerator and the denominator are $sqrtn+O(1)$, so...
– Jack D'Aurizio♦
Sep 1 at 23:33
add a comment |Â
5 Answers
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active
oldest
votes
up vote
5
down vote
accepted
HINT
By intuition we have that the $sqrt n$ terms are dominant on the others and therefore in the limit
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4sim fracsqrt nsqrt n=1$$
To check and formalize rigoursly this fact let consider
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4=fracsqrt nsqrt nfracsqrt(1+1/n)+ 3/sqrt nsqrt(1+2/n) - 4/sqrt n$$
then take the limit.
answered Sep 1 at 22:29
gimusi
70.9k73786
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up vote
2
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Using equivalents, it's very simple:
$sqrtn+1+3,:sqrtn+2-4sim_inftysqrt n$, so
$$fracsqrtn+1+3sqrtn+2-4sim_inftyfracsqrt nsqrt n=1.$$
answered Sep 1 at 22:33
Bernard
111k635102
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Just if you want to know more than the limit itself.
$$sqrtn+a=sqrt n sqrt1+frac an$$ Now, use the binomial expansion or Taylor series to get
$$sqrt1+frac an=1+fraca2 n-fraca^28 n^2+Oleft(frac1n^3right)$$
$$fracsqrtn+1 + 3sqrtn+2 - 4=fracsqrt nleft(1+frac12 n-frac18 n^2+Oleft(frac1n^3right) right)+3 sqrt nleft(1+frac1n-frac12 n^2 +Oleft(frac1n^3right)right)-4 =frac sqrt n +3 +frac1 2 sqrt n-frac1 8 n^3/2+Oleft(frac1n^5/2right) sqrt n -4 +frac1 sqrt n-frac1 2 n^3/2+Oleft(frac1n^5/2right) $$ Now, long division to get
$$fracsqrtn+1 + 3sqrtn+2 - 4=1+frac 7 sqrt n+frac 552n+Oleft(frac1n^3/2right)$$
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
add a comment |Â
up vote
1
down vote
General rules:
(1). If $lim_nto inftyA_n=A$ and $lim_nto inftyB_n=B$ then
(1a). $lim_nto infty(A_n+B_n)=A+B.$
(1b). If $Bne 0$ then $lim_nto infty (A_n/B_n)=A/B.$
(2). If $lim_nto inftyA_n=Ageq 0$ then $lim_nto infty sqrtA_n=sqrt A.$
We have $lim_nto inftyfrac sqrt n+1+3sqrt n+2-4=$ $lim_nto inftyfrac sqrt 1+1/n;+(3/sqrt n)sqrt 1+2/n ;-(4/sqrt n).$
We have $1=lim_nto infty(1+1/n)=lim_nto infty(1+2/n).$ We have $0=lim_nto infty(3/sqrt n)=lim_nto infty(-4/sqrt n).$
Applying the rules above , we obtain a limit of $1$ for both the numerator and denominator of the given expression, and therefore a limit of $1$ for the whole thing.
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
add a comment |Â
up vote
1
down vote
Based on the estimate:
$$colorbluefracsqrtn+1+3sqrtn+2<fracsqrtn+1+3sqrtn+2-4<colorredfracsqrtn+2+3sqrtn+2-4, n>14 iff \
lim_ntoinftycolorbluefracsqrtn+1+3sqrtn+2=lim_ntoinftysqrt1-frac1n+2+lim_ntoinftyfrac3sqrtn+2=1+0=1;\
lim_ntoinftycolorredfracsqrtn+2+3sqrtn+2-4=lim_ntoinftyleft(1+frac7sqrtn+2-4right)=1.$$
By the Squeeze Theorem the limit is $1$.
answered Sep 2 at 9:12
farruhota
15k2734
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
HINT
By intuition we have that the $sqrt n$ terms are dominant on the others and therefore in the limit
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4sim fracsqrt nsqrt n=1$$
To check and formalize rigoursly this fact let consider
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4=fracsqrt nsqrt nfracsqrt(1+1/n)+ 3/sqrt nsqrt(1+2/n) - 4/sqrt n$$
then take the limit.
answered Sep 1 at 22:29
gimusi
70.9k73786
add a comment |Â
up vote
5
down vote
accepted
HINT
By intuition we have that the $sqrt n$ terms are dominant on the others and therefore in the limit
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4sim fracsqrt nsqrt n=1$$
To check and formalize rigoursly this fact let consider
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4=fracsqrt nsqrt nfracsqrt(1+1/n)+ 3/sqrt nsqrt(1+2/n) - 4/sqrt n$$
then take the limit.
answered Sep 1 at 22:29
gimusi
70.9k73786
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
HINT
By intuition we have that the $sqrt n$ terms are dominant on the others and therefore in the limit
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4sim fracsqrt nsqrt n=1$$
To check and formalize rigoursly this fact let consider
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4=fracsqrt nsqrt nfracsqrt(1+1/n)+ 3/sqrt nsqrt(1+2/n) - 4/sqrt n$$
then take the limit.
answered Sep 1 at 22:29
gimusi
70.9k73786
HINT
By intuition we have that the $sqrt n$ terms are dominant on the others and therefore in the limit
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4sim fracsqrt nsqrt n=1$$
To check and formalize rigoursly this fact let consider
$$fracsqrt(n+1)+ 3sqrt(n+2) - 4=fracsqrt nsqrt nfracsqrt(1+1/n)+ 3/sqrt nsqrt(1+2/n) - 4/sqrt n$$
then take the limit.
answered Sep 1 at 22:29
gimusi
70.9k73786
edited Sep 1 at 22:35
answered Sep 1 at 22:29
gimusi
70.9k73786
answered Sep 1 at 22:29
gimusi
70.9k73786
answered Sep 1 at 22:29
gimusi
70.9k73786
70.9k73786
add a comment |Â
add a comment |Â
up vote
2
down vote
Using equivalents, it's very simple:
$sqrtn+1+3,:sqrtn+2-4sim_inftysqrt n$, so
$$fracsqrtn+1+3sqrtn+2-4sim_inftyfracsqrt nsqrt n=1.$$
answered Sep 1 at 22:33
Bernard
111k635102
add a comment |Â
up vote
2
down vote
Using equivalents, it's very simple:
$sqrtn+1+3,:sqrtn+2-4sim_inftysqrt n$, so
$$fracsqrtn+1+3sqrtn+2-4sim_inftyfracsqrt nsqrt n=1.$$
answered Sep 1 at 22:33
Bernard
111k635102
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using equivalents, it's very simple:
$sqrtn+1+3,:sqrtn+2-4sim_inftysqrt n$, so
$$fracsqrtn+1+3sqrtn+2-4sim_inftyfracsqrt nsqrt n=1.$$
answered Sep 1 at 22:33
Bernard
111k635102
Using equivalents, it's very simple:
$sqrtn+1+3,:sqrtn+2-4sim_inftysqrt n$, so
$$fracsqrtn+1+3sqrtn+2-4sim_inftyfracsqrt nsqrt n=1.$$
answered Sep 1 at 22:33
Bernard
111k635102
answered Sep 1 at 22:33
Bernard
111k635102
answered Sep 1 at 22:33
Bernard
111k635102
answered Sep 1 at 22:33
Bernard
111k635102
111k635102
add a comment |Â
add a comment |Â
up vote
2
down vote
Just if you want to know more than the limit itself.
$$sqrtn+a=sqrt n sqrt1+frac an$$ Now, use the binomial expansion or Taylor series to get
$$sqrt1+frac an=1+fraca2 n-fraca^28 n^2+Oleft(frac1n^3right)$$
$$fracsqrtn+1 + 3sqrtn+2 - 4=fracsqrt nleft(1+frac12 n-frac18 n^2+Oleft(frac1n^3right) right)+3 sqrt nleft(1+frac1n-frac12 n^2 +Oleft(frac1n^3right)right)-4 =frac sqrt n +3 +frac1 2 sqrt n-frac1 8 n^3/2+Oleft(frac1n^5/2right) sqrt n -4 +frac1 sqrt n-frac1 2 n^3/2+Oleft(frac1n^5/2right) $$ Now, long division to get
$$fracsqrtn+1 + 3sqrtn+2 - 4=1+frac 7 sqrt n+frac 552n+Oleft(frac1n^3/2right)$$
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
add a comment |Â
up vote
2
down vote
Just if you want to know more than the limit itself.
$$sqrtn+a=sqrt n sqrt1+frac an$$ Now, use the binomial expansion or Taylor series to get
$$sqrt1+frac an=1+fraca2 n-fraca^28 n^2+Oleft(frac1n^3right)$$
$$fracsqrtn+1 + 3sqrtn+2 - 4=fracsqrt nleft(1+frac12 n-frac18 n^2+Oleft(frac1n^3right) right)+3 sqrt nleft(1+frac1n-frac12 n^2 +Oleft(frac1n^3right)right)-4 =frac sqrt n +3 +frac1 2 sqrt n-frac1 8 n^3/2+Oleft(frac1n^5/2right) sqrt n -4 +frac1 sqrt n-frac1 2 n^3/2+Oleft(frac1n^5/2right) $$ Now, long division to get
$$fracsqrtn+1 + 3sqrtn+2 - 4=1+frac 7 sqrt n+frac 552n+Oleft(frac1n^3/2right)$$
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Just if you want to know more than the limit itself.
$$sqrtn+a=sqrt n sqrt1+frac an$$ Now, use the binomial expansion or Taylor series to get
$$sqrt1+frac an=1+fraca2 n-fraca^28 n^2+Oleft(frac1n^3right)$$
$$fracsqrtn+1 + 3sqrtn+2 - 4=fracsqrt nleft(1+frac12 n-frac18 n^2+Oleft(frac1n^3right) right)+3 sqrt nleft(1+frac1n-frac12 n^2 +Oleft(frac1n^3right)right)-4 =frac sqrt n +3 +frac1 2 sqrt n-frac1 8 n^3/2+Oleft(frac1n^5/2right) sqrt n -4 +frac1 sqrt n-frac1 2 n^3/2+Oleft(frac1n^5/2right) $$ Now, long division to get
$$fracsqrtn+1 + 3sqrtn+2 - 4=1+frac 7 sqrt n+frac 552n+Oleft(frac1n^3/2right)$$
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
Just if you want to know more than the limit itself.
$$sqrtn+a=sqrt n sqrt1+frac an$$ Now, use the binomial expansion or Taylor series to get
$$sqrt1+frac an=1+fraca2 n-fraca^28 n^2+Oleft(frac1n^3right)$$
$$fracsqrtn+1 + 3sqrtn+2 - 4=fracsqrt nleft(1+frac12 n-frac18 n^2+Oleft(frac1n^3right) right)+3 sqrt nleft(1+frac1n-frac12 n^2 +Oleft(frac1n^3right)right)-4 =frac sqrt n +3 +frac1 2 sqrt n-frac1 8 n^3/2+Oleft(frac1n^5/2right) sqrt n -4 +frac1 sqrt n-frac1 2 n^3/2+Oleft(frac1n^5/2right) $$ Now, long division to get
$$fracsqrtn+1 + 3sqrtn+2 - 4=1+frac 7 sqrt n+frac 552n+Oleft(frac1n^3/2right)$$
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
answered Sep 2 at 5:01
Claude Leibovici
113k1155127
113k1155127
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
add a comment |Â
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
That's a nice limit radiography!
– gimusi
Sep 2 at 7:20
add a comment |Â
up vote
1
down vote
General rules:
(1). If $lim_nto inftyA_n=A$ and $lim_nto inftyB_n=B$ then
(1a). $lim_nto infty(A_n+B_n)=A+B.$
(1b). If $Bne 0$ then $lim_nto infty (A_n/B_n)=A/B.$
(2). If $lim_nto inftyA_n=Ageq 0$ then $lim_nto infty sqrtA_n=sqrt A.$
We have $lim_nto inftyfrac sqrt n+1+3sqrt n+2-4=$ $lim_nto inftyfrac sqrt 1+1/n;+(3/sqrt n)sqrt 1+2/n ;-(4/sqrt n).$
We have $1=lim_nto infty(1+1/n)=lim_nto infty(1+2/n).$ We have $0=lim_nto infty(3/sqrt n)=lim_nto infty(-4/sqrt n).$
Applying the rules above , we obtain a limit of $1$ for both the numerator and denominator of the given expression, and therefore a limit of $1$ for the whole thing.
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
add a comment |Â
up vote
1
down vote
General rules:
(1). If $lim_nto inftyA_n=A$ and $lim_nto inftyB_n=B$ then
(1a). $lim_nto infty(A_n+B_n)=A+B.$
(1b). If $Bne 0$ then $lim_nto infty (A_n/B_n)=A/B.$
(2). If $lim_nto inftyA_n=Ageq 0$ then $lim_nto infty sqrtA_n=sqrt A.$
We have $lim_nto inftyfrac sqrt n+1+3sqrt n+2-4=$ $lim_nto inftyfrac sqrt 1+1/n;+(3/sqrt n)sqrt 1+2/n ;-(4/sqrt n).$
We have $1=lim_nto infty(1+1/n)=lim_nto infty(1+2/n).$ We have $0=lim_nto infty(3/sqrt n)=lim_nto infty(-4/sqrt n).$
Applying the rules above , we obtain a limit of $1$ for both the numerator and denominator of the given expression, and therefore a limit of $1$ for the whole thing.
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
add a comment |Â
up vote
1
down vote
up vote
1
down vote
General rules:
(1). If $lim_nto inftyA_n=A$ and $lim_nto inftyB_n=B$ then
(1a). $lim_nto infty(A_n+B_n)=A+B.$
(1b). If $Bne 0$ then $lim_nto infty (A_n/B_n)=A/B.$
(2). If $lim_nto inftyA_n=Ageq 0$ then $lim_nto infty sqrtA_n=sqrt A.$
We have $lim_nto inftyfrac sqrt n+1+3sqrt n+2-4=$ $lim_nto inftyfrac sqrt 1+1/n;+(3/sqrt n)sqrt 1+2/n ;-(4/sqrt n).$
We have $1=lim_nto infty(1+1/n)=lim_nto infty(1+2/n).$ We have $0=lim_nto infty(3/sqrt n)=lim_nto infty(-4/sqrt n).$
Applying the rules above , we obtain a limit of $1$ for both the numerator and denominator of the given expression, and therefore a limit of $1$ for the whole thing.
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
General rules:
(1). If $lim_nto inftyA_n=A$ and $lim_nto inftyB_n=B$ then
(1a). $lim_nto infty(A_n+B_n)=A+B.$
(1b). If $Bne 0$ then $lim_nto infty (A_n/B_n)=A/B.$
(2). If $lim_nto inftyA_n=Ageq 0$ then $lim_nto infty sqrtA_n=sqrt A.$
We have $lim_nto inftyfrac sqrt n+1+3sqrt n+2-4=$ $lim_nto inftyfrac sqrt 1+1/n;+(3/sqrt n)sqrt 1+2/n ;-(4/sqrt n).$
We have $1=lim_nto infty(1+1/n)=lim_nto infty(1+2/n).$ We have $0=lim_nto infty(3/sqrt n)=lim_nto infty(-4/sqrt n).$
Applying the rules above , we obtain a limit of $1$ for both the numerator and denominator of the given expression, and therefore a limit of $1$ for the whole thing.
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
answered Sep 2 at 7:37
DanielWainfleet
32.1k31644
32.1k31644
add a comment |Â
add a comment |Â
up vote
1
down vote
Based on the estimate:
$$colorbluefracsqrtn+1+3sqrtn+2<fracsqrtn+1+3sqrtn+2-4<colorredfracsqrtn+2+3sqrtn+2-4, n>14 iff \
lim_ntoinftycolorbluefracsqrtn+1+3sqrtn+2=lim_ntoinftysqrt1-frac1n+2+lim_ntoinftyfrac3sqrtn+2=1+0=1;\
lim_ntoinftycolorredfracsqrtn+2+3sqrtn+2-4=lim_ntoinftyleft(1+frac7sqrtn+2-4right)=1.$$
By the Squeeze Theorem the limit is $1$.
answered Sep 2 at 9:12
farruhota
15k2734
add a comment |Â
up vote
1
down vote
Based on the estimate:
$$colorbluefracsqrtn+1+3sqrtn+2<fracsqrtn+1+3sqrtn+2-4<colorredfracsqrtn+2+3sqrtn+2-4, n>14 iff \
lim_ntoinftycolorbluefracsqrtn+1+3sqrtn+2=lim_ntoinftysqrt1-frac1n+2+lim_ntoinftyfrac3sqrtn+2=1+0=1;\
lim_ntoinftycolorredfracsqrtn+2+3sqrtn+2-4=lim_ntoinftyleft(1+frac7sqrtn+2-4right)=1.$$
By the Squeeze Theorem the limit is $1$.
answered Sep 2 at 9:12
farruhota
15k2734
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Based on the estimate:
$$colorbluefracsqrtn+1+3sqrtn+2<fracsqrtn+1+3sqrtn+2-4<colorredfracsqrtn+2+3sqrtn+2-4, n>14 iff \
lim_ntoinftycolorbluefracsqrtn+1+3sqrtn+2=lim_ntoinftysqrt1-frac1n+2+lim_ntoinftyfrac3sqrtn+2=1+0=1;\
lim_ntoinftycolorredfracsqrtn+2+3sqrtn+2-4=lim_ntoinftyleft(1+frac7sqrtn+2-4right)=1.$$
By the Squeeze Theorem the limit is $1$.
answered Sep 2 at 9:12
farruhota
15k2734
Based on the estimate:
$$colorbluefracsqrtn+1+3sqrtn+2<fracsqrtn+1+3sqrtn+2-4<colorredfracsqrtn+2+3sqrtn+2-4, n>14 iff \
lim_ntoinftycolorbluefracsqrtn+1+3sqrtn+2=lim_ntoinftysqrt1-frac1n+2+lim_ntoinftyfrac3sqrtn+2=1+0=1;\
lim_ntoinftycolorredfracsqrtn+2+3sqrtn+2-4=lim_ntoinftyleft(1+frac7sqrtn+2-4right)=1.$$
By the Squeeze Theorem the limit is $1$.
answered Sep 2 at 9:12
farruhota
15k2734
answered Sep 2 at 9:12
farruhota
15k2734
answered Sep 2 at 9:12
farruhota
15k2734
answered Sep 2 at 9:12
farruhota
15k2734
15k2734
add a comment |Â
add a comment |Â
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1
Both the numerator and the denominator are $sqrtn+O(1)$, so...
– Jack D'Aurizio♦
Sep 1 at 23:33