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Shorter Way to Solve $lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$
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up vote
3
down vote
favorite
The limit to find is
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$lim_xto 0frac(x-ln(1+x))(x^2+xln(x+1)+ln^2(x+1))(sin x-x)(sin x+x)$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?
limits
edited Aug 2 at 15:43
Jam
4,10111230
asked Aug 2 at 15:26
NotADeveloper
5321210
add a comment |Â
up vote
3
down vote
favorite
The limit to find is
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$lim_xto 0frac(x-ln(1+x))(x^2+xln(x+1)+ln^2(x+1))(sin x-x)(sin x+x)$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?
limits
edited Aug 2 at 15:43
Jam
4,10111230
asked Aug 2 at 15:26
NotADeveloper
5321210
The Limit should be $$-frac92$$
– Dr. Sonnhard Graubner
Aug 2 at 15:36
3
I've changed the title to reflect that you've solved it but want a shorter way to do so
– Jam
Aug 2 at 15:43
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The limit to find is
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$lim_xto 0frac(x-ln(1+x))(x^2+xln(x+1)+ln^2(x+1))(sin x-x)(sin x+x)$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?
limits
edited Aug 2 at 15:43
Jam
4,10111230
asked Aug 2 at 15:26
NotADeveloper
5321210
The limit to find is
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$lim_xto 0frac(x-ln(1+x))(x^2+xln(x+1)+ln^2(x+1))(sin x-x)(sin x+x)$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?
limits
edited Aug 2 at 15:43
Jam
4,10111230
asked Aug 2 at 15:26
NotADeveloper
5321210
edited Aug 2 at 15:43
Jam
4,10111230
edited Aug 2 at 15:43
Jam
4,10111230
edited Aug 2 at 15:43
Jam
4,10111230
4,10111230
asked Aug 2 at 15:26
NotADeveloper
5321210
asked Aug 2 at 15:26
NotADeveloper
5321210
asked Aug 2 at 15:26
NotADeveloper
5321210
5321210
The Limit should be $$-frac92$$
– Dr. Sonnhard Graubner
Aug 2 at 15:36
3
I've changed the title to reflect that you've solved it but want a shorter way to do so
– Jam
Aug 2 at 15:43
add a comment |Â
The Limit should be $$-frac92$$
– Dr. Sonnhard Graubner
Aug 2 at 15:36
3
I've changed the title to reflect that you've solved it but want a shorter way to do so
– Jam
Aug 2 at 15:43
The Limit should be $$-frac92$$
– Dr. Sonnhard Graubner
Aug 2 at 15:36
The Limit should be $$-frac92$$
– Dr. Sonnhard Graubner
Aug 2 at 15:36
3
3
I've changed the title to reflect that you've solved it but want a shorter way to do so
– Jam
Aug 2 at 15:43
I've changed the title to reflect that you've solved it but want a shorter way to do so
– Jam
Aug 2 at 15:43
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
2
down vote
accepted
Taylor Series around $0$ of $ln(1+x)$:
$$ln(1+x) = x - fracx^22 + O(x^3)$$
$$ln^3(1+x) = x^3 -frac32x^4 + O(x^5)$$
Taylor Series around $0$ of $sin(x)$:
$$sin(x) = x - fracx^36 + O(x^5)$$
$$sin^2(x) = x^2 - fracx^43 + O(x^6)$$
So
$$fracx^3 - ln^3(1+x) sin^2(x) - x^2 simeq fracx^3 - x^3 +frac32x^4x^2 - fracx^43 - x^2 = fracfrac32x^4- fracx^43 = - frac92$$
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
add a comment |Â
up vote
3
down vote
Divide numerator and denominator by $x^4$ and then the denominator is handled as $$fracsin x-xx^3cdotfracsin x +x x to (-1/6)(2)=-frac13$$ The numerator on the other hand is handled as $$fracx-log(1+x)x^2cdotfrac x^2+xlog(1+x)+log^2(1+x) x^2 tofrac12cdot(1+1+1)=frac32$$ and hence the desired limit is $-9/2$.
answered Aug 2 at 17:00
Paramanand Singh
45k553142
add a comment |Â
up vote
2
down vote
You could use Taylor series at $x=0$ to compute the limit. Commonly used Taylor Series
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
By using Taylor series we get
$$-frac92+frac21x4-frac249x^240+frac13x^32-.....O(x^6)$$at $x=0$ we get
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2=-frac92$$
answered Aug 2 at 15:48
Key Flex
3,692422
add a comment |Â
up vote
2
down vote
Using $sin(x)=x-fracx^36+mathcalO(x^5)$ and $ln(1+x)=x-fracx^22+mathcalO(x^3)$:
$$beginalignedfracx^3-ln^3(1+x)sin^2(x)-x^2
&=fracx^3-left(x-fracx^22+mathcalO(x^3)right)^3left(x-fracx^36+mathcalO(x^5)right)^2-x^2\
&=fracx^3-left(x^3- frac3 x^42+mathcalO(x^5)right)left(x^2-fracx^43+mathcalO(x^6)right)-x^2\
&=fracfrac3 x^42+mathcalO(x^5)-fracx^43+mathcalO(x^6)
endaligned$$
answered Aug 2 at 15:57
Jam
4,10111230
add a comment |Â
up vote
1
down vote
Hint: By a series Expansion we get
$-frac92+frac214x-frac24940x^2+frac132x^3-frac186192800x^4+frac336475040x^5+...$
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Taylor Series around $0$ of $ln(1+x)$:
$$ln(1+x) = x - fracx^22 + O(x^3)$$
$$ln^3(1+x) = x^3 -frac32x^4 + O(x^5)$$
Taylor Series around $0$ of $sin(x)$:
$$sin(x) = x - fracx^36 + O(x^5)$$
$$sin^2(x) = x^2 - fracx^43 + O(x^6)$$
So
$$fracx^3 - ln^3(1+x) sin^2(x) - x^2 simeq fracx^3 - x^3 +frac32x^4x^2 - fracx^43 - x^2 = fracfrac32x^4- fracx^43 = - frac92$$
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
add a comment |Â
up vote
2
down vote
accepted
Taylor Series around $0$ of $ln(1+x)$:
$$ln(1+x) = x - fracx^22 + O(x^3)$$
$$ln^3(1+x) = x^3 -frac32x^4 + O(x^5)$$
Taylor Series around $0$ of $sin(x)$:
$$sin(x) = x - fracx^36 + O(x^5)$$
$$sin^2(x) = x^2 - fracx^43 + O(x^6)$$
So
$$fracx^3 - ln^3(1+x) sin^2(x) - x^2 simeq fracx^3 - x^3 +frac32x^4x^2 - fracx^43 - x^2 = fracfrac32x^4- fracx^43 = - frac92$$
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Taylor Series around $0$ of $ln(1+x)$:
$$ln(1+x) = x - fracx^22 + O(x^3)$$
$$ln^3(1+x) = x^3 -frac32x^4 + O(x^5)$$
Taylor Series around $0$ of $sin(x)$:
$$sin(x) = x - fracx^36 + O(x^5)$$
$$sin^2(x) = x^2 - fracx^43 + O(x^6)$$
So
$$fracx^3 - ln^3(1+x) sin^2(x) - x^2 simeq fracx^3 - x^3 +frac32x^4x^2 - fracx^43 - x^2 = fracfrac32x^4- fracx^43 = - frac92$$
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
Taylor Series around $0$ of $ln(1+x)$:
$$ln(1+x) = x - fracx^22 + O(x^3)$$
$$ln^3(1+x) = x^3 -frac32x^4 + O(x^5)$$
Taylor Series around $0$ of $sin(x)$:
$$sin(x) = x - fracx^36 + O(x^5)$$
$$sin^2(x) = x^2 - fracx^43 + O(x^6)$$
So
$$fracx^3 - ln^3(1+x) sin^2(x) - x^2 simeq fracx^3 - x^3 +frac32x^4x^2 - fracx^43 - x^2 = fracfrac32x^4- fracx^43 = - frac92$$
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
answered Aug 2 at 15:50
Ahmad Bazzi
2,172417
2,172417
add a comment |Â
add a comment |Â
up vote
3
down vote
Divide numerator and denominator by $x^4$ and then the denominator is handled as $$fracsin x-xx^3cdotfracsin x +x x to (-1/6)(2)=-frac13$$ The numerator on the other hand is handled as $$fracx-log(1+x)x^2cdotfrac x^2+xlog(1+x)+log^2(1+x) x^2 tofrac12cdot(1+1+1)=frac32$$ and hence the desired limit is $-9/2$.
answered Aug 2 at 17:00
Paramanand Singh
45k553142
add a comment |Â
up vote
3
down vote
Divide numerator and denominator by $x^4$ and then the denominator is handled as $$fracsin x-xx^3cdotfracsin x +x x to (-1/6)(2)=-frac13$$ The numerator on the other hand is handled as $$fracx-log(1+x)x^2cdotfrac x^2+xlog(1+x)+log^2(1+x) x^2 tofrac12cdot(1+1+1)=frac32$$ and hence the desired limit is $-9/2$.
answered Aug 2 at 17:00
Paramanand Singh
45k553142
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Divide numerator and denominator by $x^4$ and then the denominator is handled as $$fracsin x-xx^3cdotfracsin x +x x to (-1/6)(2)=-frac13$$ The numerator on the other hand is handled as $$fracx-log(1+x)x^2cdotfrac x^2+xlog(1+x)+log^2(1+x) x^2 tofrac12cdot(1+1+1)=frac32$$ and hence the desired limit is $-9/2$.
answered Aug 2 at 17:00
Paramanand Singh
45k553142
Divide numerator and denominator by $x^4$ and then the denominator is handled as $$fracsin x-xx^3cdotfracsin x +x x to (-1/6)(2)=-frac13$$ The numerator on the other hand is handled as $$fracx-log(1+x)x^2cdotfrac x^2+xlog(1+x)+log^2(1+x) x^2 tofrac12cdot(1+1+1)=frac32$$ and hence the desired limit is $-9/2$.
answered Aug 2 at 17:00
Paramanand Singh
45k553142
answered Aug 2 at 17:00
Paramanand Singh
45k553142
answered Aug 2 at 17:00
Paramanand Singh
45k553142
answered Aug 2 at 17:00
Paramanand Singh
45k553142
45k553142
add a comment |Â
add a comment |Â
up vote
2
down vote
You could use Taylor series at $x=0$ to compute the limit. Commonly used Taylor Series
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
By using Taylor series we get
$$-frac92+frac21x4-frac249x^240+frac13x^32-.....O(x^6)$$at $x=0$ we get
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2=-frac92$$
answered Aug 2 at 15:48
Key Flex
3,692422
add a comment |Â
up vote
2
down vote
You could use Taylor series at $x=0$ to compute the limit. Commonly used Taylor Series
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
By using Taylor series we get
$$-frac92+frac21x4-frac249x^240+frac13x^32-.....O(x^6)$$at $x=0$ we get
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2=-frac92$$
answered Aug 2 at 15:48
Key Flex
3,692422
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You could use Taylor series at $x=0$ to compute the limit. Commonly used Taylor Series
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
By using Taylor series we get
$$-frac92+frac21x4-frac249x^240+frac13x^32-.....O(x^6)$$at $x=0$ we get
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2=-frac92$$
answered Aug 2 at 15:48
Key Flex
3,692422
You could use Taylor series at $x=0$ to compute the limit. Commonly used Taylor Series
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2$$
By using Taylor series we get
$$-frac92+frac21x4-frac249x^240+frac13x^32-.....O(x^6)$$at $x=0$ we get
$$lim_xto 0 fracx^3-ln^3(1+x)sin^2x-x^2=-frac92$$
answered Aug 2 at 15:48
Key Flex
3,692422
answered Aug 2 at 15:48
Key Flex
3,692422
answered Aug 2 at 15:48
Key Flex
3,692422
answered Aug 2 at 15:48
Key Flex
3,692422
3,692422
add a comment |Â
add a comment |Â
up vote
2
down vote
Using $sin(x)=x-fracx^36+mathcalO(x^5)$ and $ln(1+x)=x-fracx^22+mathcalO(x^3)$:
$$beginalignedfracx^3-ln^3(1+x)sin^2(x)-x^2
&=fracx^3-left(x-fracx^22+mathcalO(x^3)right)^3left(x-fracx^36+mathcalO(x^5)right)^2-x^2\
&=fracx^3-left(x^3- frac3 x^42+mathcalO(x^5)right)left(x^2-fracx^43+mathcalO(x^6)right)-x^2\
&=fracfrac3 x^42+mathcalO(x^5)-fracx^43+mathcalO(x^6)
endaligned$$
answered Aug 2 at 15:57
Jam
4,10111230
add a comment |Â
up vote
2
down vote
Using $sin(x)=x-fracx^36+mathcalO(x^5)$ and $ln(1+x)=x-fracx^22+mathcalO(x^3)$:
$$beginalignedfracx^3-ln^3(1+x)sin^2(x)-x^2
&=fracx^3-left(x-fracx^22+mathcalO(x^3)right)^3left(x-fracx^36+mathcalO(x^5)right)^2-x^2\
&=fracx^3-left(x^3- frac3 x^42+mathcalO(x^5)right)left(x^2-fracx^43+mathcalO(x^6)right)-x^2\
&=fracfrac3 x^42+mathcalO(x^5)-fracx^43+mathcalO(x^6)
endaligned$$
answered Aug 2 at 15:57
Jam
4,10111230
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using $sin(x)=x-fracx^36+mathcalO(x^5)$ and $ln(1+x)=x-fracx^22+mathcalO(x^3)$:
$$beginalignedfracx^3-ln^3(1+x)sin^2(x)-x^2
&=fracx^3-left(x-fracx^22+mathcalO(x^3)right)^3left(x-fracx^36+mathcalO(x^5)right)^2-x^2\
&=fracx^3-left(x^3- frac3 x^42+mathcalO(x^5)right)left(x^2-fracx^43+mathcalO(x^6)right)-x^2\
&=fracfrac3 x^42+mathcalO(x^5)-fracx^43+mathcalO(x^6)
endaligned$$
answered Aug 2 at 15:57
Jam
4,10111230
Using $sin(x)=x-fracx^36+mathcalO(x^5)$ and $ln(1+x)=x-fracx^22+mathcalO(x^3)$:
$$beginalignedfracx^3-ln^3(1+x)sin^2(x)-x^2
&=fracx^3-left(x-fracx^22+mathcalO(x^3)right)^3left(x-fracx^36+mathcalO(x^5)right)^2-x^2\
&=fracx^3-left(x^3- frac3 x^42+mathcalO(x^5)right)left(x^2-fracx^43+mathcalO(x^6)right)-x^2\
&=fracfrac3 x^42+mathcalO(x^5)-fracx^43+mathcalO(x^6)
endaligned$$
answered Aug 2 at 15:57
Jam
4,10111230
edited Aug 2 at 16:13
answered Aug 2 at 15:57
Jam
4,10111230
answered Aug 2 at 15:57
Jam
4,10111230
answered Aug 2 at 15:57
Jam
4,10111230
4,10111230
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint: By a series Expansion we get
$-frac92+frac214x-frac24940x^2+frac132x^3-frac186192800x^4+frac336475040x^5+...$
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
up vote
1
down vote
Hint: By a series Expansion we get
$-frac92+frac214x-frac24940x^2+frac132x^3-frac186192800x^4+frac336475040x^5+...$
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: By a series Expansion we get
$-frac92+frac214x-frac24940x^2+frac132x^3-frac186192800x^4+frac336475040x^5+...$
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
Hint: By a series Expansion we get
$-frac92+frac214x-frac24940x^2+frac132x^3-frac186192800x^4+frac336475040x^5+...$
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
answered Aug 2 at 15:48
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
add a comment |Â
add a comment |Â
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The Limit should be $$-frac92$$
– Dr. Sonnhard Graubner
Aug 2 at 15:36
3
I've changed the title to reflect that you've solved it but want a shorter way to do so
– Jam
Aug 2 at 15:43