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Evaluating $lim_xto0 fraccos x - cos 3xsin 3x^2 - sin x^2$
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$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -lim_xto 0fraccos x-cos(3x)sin(3x^2)cdotfrac11-dfracsin(3x^2)sin(x^2) $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
calculus limits trigonometry
edited Jul 31 at 2:37
user 108128
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asked Jul 30 at 20:14
Pavel Iljiev
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up vote
9
down vote
favorite
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -lim_xto 0fraccos x-cos(3x)sin(3x^2)cdotfrac11-dfracsin(3x^2)sin(x^2) $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
calculus limits trigonometry
edited Jul 31 at 2:37
user 108128
18.9k41544
asked Jul 30 at 20:14
Pavel Iljiev
754
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -lim_xto 0fraccos x-cos(3x)sin(3x^2)cdotfrac11-dfracsin(3x^2)sin(x^2) $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
calculus limits trigonometry
edited Jul 31 at 2:37
user 108128
18.9k41544
asked Jul 30 at 20:14
Pavel Iljiev
754
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -lim_xto 0fraccos x-cos(3x)sin(3x^2)cdotfrac11-dfracsin(3x^2)sin(x^2) $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
calculus limits trigonometry
edited Jul 31 at 2:37
user 108128
18.9k41544
asked Jul 30 at 20:14
Pavel Iljiev
754
edited Jul 31 at 2:37
user 108128
18.9k41544
edited Jul 31 at 2:37
user 108128
18.9k41544
edited Jul 31 at 2:37
user 108128
18.9k41544
18.9k41544
asked Jul 30 at 20:14
Pavel Iljiev
754
asked Jul 30 at 20:14
Pavel Iljiev
754
asked Jul 30 at 20:14
Pavel Iljiev
754
754
add a comment |Â
add a comment |Â
8 Answers
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up vote
12
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accepted
By standard limits
$fracsin xx to 1$
$frac1-cos xx^2 to frac12$
we have that
$$fraccos x-cos(3x)sin(3x^2)-sin(x^2)=fracfraccos x-1+1- cos(3x)x^2 fracsin(3x^2)-sin(x^2)x^2=frac-frac1-cos xx^2+9frac1- cos(3x)(3x)^2 3fracsin(3x^2)3x^2-fracsin(x^2)x^2tofrac-frac12+frac923-1=2$$
answered Jul 30 at 20:22
gimusi
64.2k73480
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up vote
8
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Hint: Use
$$sin 3a=3sin a-4sin^3a$$
$$cos 3a=4cos^3a-3cos a$$
Edit:
After substutution it is
$$lim_xto0frac2cos xsin^2xsin x^2cos2x^2=2$$
answered Jul 30 at 20:21
user 108128
18.9k41544
1
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
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I would use Taylor polynomials at order $2$.
$$fraccos x-cos (3x)sin (3x^2)-sin (x^2)=frac1-fracx^22-left(1-frac(3x)^22right)+o(x^2)3x^2-x^2+o(x^2)=frac4x^2+o(x^2)2x^2+o(x^2)=2+o(1)
$$
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
5
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
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7
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Hint: Use the factorisation formula $$cos a -cos b =-2sina+bover 2sina-bover 2$$
and
$$sin a -sin b =2sina-bover 2cosa+bover 2$$
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) = lim_xto0frac2colorredsin 2xcdot colorgreensin xcdotcolorbluex^2colorbluesin (x^2)cos (2x^2)cdotcolorred2xcdot colorgreenx =2$$
answered Jul 30 at 20:17
greedoid
26k93473
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
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up vote
4
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Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$cos u=1-fracu^22+o(u^2),qquad sin u=u+o(u)$$
so
beginalign
cos x-cos 3x&=1-fracx^22+o(x^2)-Bigl(1-frac9x^22+o(x^2)Bigr)= 4x^2+o(x^2)\
sin 3x^2-sin x^2&=3x^2+o(x^2)-bigl(sin x^2+o(x^2)bigr)=2x^2+o(x^2).
endalign
Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence
$$fraccos x-cos 3xsin 3x^2-sin x^2sim_0frac4x^22x^2=2.$$
answered Jul 30 at 20:30
Bernard
110k635102
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2
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Use the Maclaurin series
$$cos x = 1-1/2 ; x^2+1/24 ; x^4 +O(x^6)$$
$$cos 3x = 1-9/2 ; x^2+27/8 ; x^4 +O(x^6)$$
$$sin x^2 = x^2-1/6; x^6+O(x^8)$$
$$sin 3x^2 = 3x^2-9/2 ; x^6+O(x^8)$$
then quotient is
$$frac-1/2 + 9/23-1 + O(x^2)$$
and therefore the limit is 2.
answered Jul 30 at 20:30
gammatester
15.7k21429
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
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2
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Using l'Hôpital directly is very awkward.
The denominator has $x^2$, but it is immediate that
$$
lim_tto0fracsin3t-sin tt=3-1=2
$$
Consequently, also
$$
lim_xto0fracsin(3x^2)-sin(x^2)2x^2=1
$$
Good! Now we can write our limit in the form
$$
lim_xto0fraccos x-cos 3x2x^2frac2x^2sin(3x^2)-sin(x^2)
$$
The first fraction can be easily dealt with:
$$
lim_xto0fraccos x-cos 3x2x^2=
lim_xto0frac3sin3x-sin x4x=
lim_xto0frac9cos3x-cos x4=2
$$
For completeness, a much simpler strategy is using Taylor expansion:
$$
fraccos x-cos3xsin(3x^2)-sin(x^2)=
frac1-x^2/2-1+(3x)^2/2+o(x^2)3x^2-x^2+o(x^2)=
frac4+o(1)2+o(1)
$$
so the limit is $2$.
answered Jul 30 at 21:35
egreg
164k1180187
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
1
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
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$$
cos x-cos(3x) = cos xleft(sin^2x+3sin^2 xright)
$$
then
$$
fraccos x-cos (3x)sin (3x^2)-sin (x^2) = fracx^2x^2cos xleft(fracsin^2x+3sin^2 xsin (3x^2)-sin (x^2)right)=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right)
$$
hence
$$
lim_xto 0=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right) = 1cdotleft(frac1+33-1right) = 2
$$
answered Jul 30 at 22:01
Cesareo
5,5912412
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
By standard limits
$fracsin xx to 1$
$frac1-cos xx^2 to frac12$
we have that
$$fraccos x-cos(3x)sin(3x^2)-sin(x^2)=fracfraccos x-1+1- cos(3x)x^2 fracsin(3x^2)-sin(x^2)x^2=frac-frac1-cos xx^2+9frac1- cos(3x)(3x)^2 3fracsin(3x^2)3x^2-fracsin(x^2)x^2tofrac-frac12+frac923-1=2$$
answered Jul 30 at 20:22
gimusi
64.2k73480
add a comment |Â
up vote
12
down vote
accepted
By standard limits
$fracsin xx to 1$
$frac1-cos xx^2 to frac12$
we have that
$$fraccos x-cos(3x)sin(3x^2)-sin(x^2)=fracfraccos x-1+1- cos(3x)x^2 fracsin(3x^2)-sin(x^2)x^2=frac-frac1-cos xx^2+9frac1- cos(3x)(3x)^2 3fracsin(3x^2)3x^2-fracsin(x^2)x^2tofrac-frac12+frac923-1=2$$
answered Jul 30 at 20:22
gimusi
64.2k73480
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
By standard limits
$fracsin xx to 1$
$frac1-cos xx^2 to frac12$
we have that
$$fraccos x-cos(3x)sin(3x^2)-sin(x^2)=fracfraccos x-1+1- cos(3x)x^2 fracsin(3x^2)-sin(x^2)x^2=frac-frac1-cos xx^2+9frac1- cos(3x)(3x)^2 3fracsin(3x^2)3x^2-fracsin(x^2)x^2tofrac-frac12+frac923-1=2$$
answered Jul 30 at 20:22
gimusi
64.2k73480
By standard limits
$fracsin xx to 1$
$frac1-cos xx^2 to frac12$
we have that
$$fraccos x-cos(3x)sin(3x^2)-sin(x^2)=fracfraccos x-1+1- cos(3x)x^2 fracsin(3x^2)-sin(x^2)x^2=frac-frac1-cos xx^2+9frac1- cos(3x)(3x)^2 3fracsin(3x^2)3x^2-fracsin(x^2)x^2tofrac-frac12+frac923-1=2$$
answered Jul 30 at 20:22
gimusi
64.2k73480
edited Jul 30 at 20:50
answered Jul 30 at 20:22
gimusi
64.2k73480
answered Jul 30 at 20:22
gimusi
64.2k73480
answered Jul 30 at 20:22
gimusi
64.2k73480
64.2k73480
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up vote
8
down vote
Hint: Use
$$sin 3a=3sin a-4sin^3a$$
$$cos 3a=4cos^3a-3cos a$$
Edit:
After substutution it is
$$lim_xto0frac2cos xsin^2xsin x^2cos2x^2=2$$
answered Jul 30 at 20:21
user 108128
18.9k41544
1
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
add a comment |Â
up vote
8
down vote
Hint: Use
$$sin 3a=3sin a-4sin^3a$$
$$cos 3a=4cos^3a-3cos a$$
Edit:
After substutution it is
$$lim_xto0frac2cos xsin^2xsin x^2cos2x^2=2$$
answered Jul 30 at 20:21
user 108128
18.9k41544
1
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Hint: Use
$$sin 3a=3sin a-4sin^3a$$
$$cos 3a=4cos^3a-3cos a$$
Edit:
After substutution it is
$$lim_xto0frac2cos xsin^2xsin x^2cos2x^2=2$$
answered Jul 30 at 20:21
user 108128
18.9k41544
Hint: Use
$$sin 3a=3sin a-4sin^3a$$
$$cos 3a=4cos^3a-3cos a$$
Edit:
After substutution it is
$$lim_xto0frac2cos xsin^2xsin x^2cos2x^2=2$$
answered Jul 30 at 20:21
user 108128
18.9k41544
edited Jul 30 at 20:29
answered Jul 30 at 20:21
user 108128
18.9k41544
answered Jul 30 at 20:21
user 108128
18.9k41544
answered Jul 30 at 20:21
user 108128
18.9k41544
18.9k41544
1
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
add a comment |Â
1
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
1
1
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
Now the votes are balanced!
– user 108128
Jul 30 at 21:01
add a comment |Â
up vote
8
down vote
I would use Taylor polynomials at order $2$.
$$fraccos x-cos (3x)sin (3x^2)-sin (x^2)=frac1-fracx^22-left(1-frac(3x)^22right)+o(x^2)3x^2-x^2+o(x^2)=frac4x^2+o(x^2)2x^2+o(x^2)=2+o(1)
$$
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
5
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
add a comment |Â
up vote
8
down vote
I would use Taylor polynomials at order $2$.
$$fraccos x-cos (3x)sin (3x^2)-sin (x^2)=frac1-fracx^22-left(1-frac(3x)^22right)+o(x^2)3x^2-x^2+o(x^2)=frac4x^2+o(x^2)2x^2+o(x^2)=2+o(1)
$$
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
5
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
add a comment |Â
up vote
8
down vote
up vote
8
down vote
I would use Taylor polynomials at order $2$.
$$fraccos x-cos (3x)sin (3x^2)-sin (x^2)=frac1-fracx^22-left(1-frac(3x)^22right)+o(x^2)3x^2-x^2+o(x^2)=frac4x^2+o(x^2)2x^2+o(x^2)=2+o(1)
$$
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
I would use Taylor polynomials at order $2$.
$$fraccos x-cos (3x)sin (3x^2)-sin (x^2)=frac1-fracx^22-left(1-frac(3x)^22right)+o(x^2)3x^2-x^2+o(x^2)=frac4x^2+o(x^2)2x^2+o(x^2)=2+o(1)
$$
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
edited Jul 31 at 10:39
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
answered Jul 30 at 20:31
mathcounterexamples.net
23.1k21651
23.1k21651
5
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
add a comment |Â
5
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
5
5
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent.
– AccidentalFourierTransform
Jul 30 at 20:55
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected.
– mathcounterexamples.net
Jul 31 at 10:41
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
@mathcounterexamples.net Well done! You are welcome, Bye.
– gimusi
Jul 31 at 13:45
add a comment |Â
up vote
7
down vote
Hint: Use the factorisation formula $$cos a -cos b =-2sina+bover 2sina-bover 2$$
and
$$sin a -sin b =2sina-bover 2cosa+bover 2$$
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) = lim_xto0frac2colorredsin 2xcdot colorgreensin xcdotcolorbluex^2colorbluesin (x^2)cos (2x^2)cdotcolorred2xcdot colorgreenx =2$$
answered Jul 30 at 20:17
greedoid
26k93473
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
add a comment |Â
up vote
7
down vote
Hint: Use the factorisation formula $$cos a -cos b =-2sina+bover 2sina-bover 2$$
and
$$sin a -sin b =2sina-bover 2cosa+bover 2$$
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) = lim_xto0frac2colorredsin 2xcdot colorgreensin xcdotcolorbluex^2colorbluesin (x^2)cos (2x^2)cdotcolorred2xcdot colorgreenx =2$$
answered Jul 30 at 20:17
greedoid
26k93473
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Hint: Use the factorisation formula $$cos a -cos b =-2sina+bover 2sina-bover 2$$
and
$$sin a -sin b =2sina-bover 2cosa+bover 2$$
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) = lim_xto0frac2colorredsin 2xcdot colorgreensin xcdotcolorbluex^2colorbluesin (x^2)cos (2x^2)cdotcolorred2xcdot colorgreenx =2$$
answered Jul 30 at 20:17
greedoid
26k93473
Hint: Use the factorisation formula $$cos a -cos b =-2sina+bover 2sina-bover 2$$
and
$$sin a -sin b =2sina-bover 2cosa+bover 2$$
$$ lim_xto0fraccos x-cos (3x)sin (3x^2)-sin (x^2) = lim_xto0frac2colorredsin 2xcdot colorgreensin xcdotcolorbluex^2colorbluesin (x^2)cos (2x^2)cdotcolorred2xcdot colorgreenx =2$$
answered Jul 30 at 20:17
greedoid
26k93473
edited Jul 30 at 20:28
answered Jul 30 at 20:17
greedoid
26k93473
answered Jul 30 at 20:17
greedoid
26k93473
answered Jul 30 at 20:17
greedoid
26k93473
26k93473
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
add a comment |Â
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
This is my way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Jul 31 at 7:34
add a comment |Â
up vote
4
down vote
Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$cos u=1-fracu^22+o(u^2),qquad sin u=u+o(u)$$
so
beginalign
cos x-cos 3x&=1-fracx^22+o(x^2)-Bigl(1-frac9x^22+o(x^2)Bigr)= 4x^2+o(x^2)\
sin 3x^2-sin x^2&=3x^2+o(x^2)-bigl(sin x^2+o(x^2)bigr)=2x^2+o(x^2).
endalign
Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence
$$fraccos x-cos 3xsin 3x^2-sin x^2sim_0frac4x^22x^2=2.$$
answered Jul 30 at 20:30
Bernard
110k635102
add a comment |Â
up vote
4
down vote
Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$cos u=1-fracu^22+o(u^2),qquad sin u=u+o(u)$$
so
beginalign
cos x-cos 3x&=1-fracx^22+o(x^2)-Bigl(1-frac9x^22+o(x^2)Bigr)= 4x^2+o(x^2)\
sin 3x^2-sin x^2&=3x^2+o(x^2)-bigl(sin x^2+o(x^2)bigr)=2x^2+o(x^2).
endalign
Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence
$$fraccos x-cos 3xsin 3x^2-sin x^2sim_0frac4x^22x^2=2.$$
answered Jul 30 at 20:30
Bernard
110k635102
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$cos u=1-fracu^22+o(u^2),qquad sin u=u+o(u)$$
so
beginalign
cos x-cos 3x&=1-fracx^22+o(x^2)-Bigl(1-frac9x^22+o(x^2)Bigr)= 4x^2+o(x^2)\
sin 3x^2-sin x^2&=3x^2+o(x^2)-bigl(sin x^2+o(x^2)bigr)=2x^2+o(x^2).
endalign
Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence
$$fraccos x-cos 3xsin 3x^2-sin x^2sim_0frac4x^22x^2=2.$$
answered Jul 30 at 20:30
Bernard
110k635102
Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$cos u=1-fracu^22+o(u^2),qquad sin u=u+o(u)$$
so
beginalign
cos x-cos 3x&=1-fracx^22+o(x^2)-Bigl(1-frac9x^22+o(x^2)Bigr)= 4x^2+o(x^2)\
sin 3x^2-sin x^2&=3x^2+o(x^2)-bigl(sin x^2+o(x^2)bigr)=2x^2+o(x^2).
endalign
Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence
$$fraccos x-cos 3xsin 3x^2-sin x^2sim_0frac4x^22x^2=2.$$
answered Jul 30 at 20:30
Bernard
110k635102
answered Jul 30 at 20:30
Bernard
110k635102
answered Jul 30 at 20:30
Bernard
110k635102
answered Jul 30 at 20:30
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
up vote
2
down vote
Use the Maclaurin series
$$cos x = 1-1/2 ; x^2+1/24 ; x^4 +O(x^6)$$
$$cos 3x = 1-9/2 ; x^2+27/8 ; x^4 +O(x^6)$$
$$sin x^2 = x^2-1/6; x^6+O(x^8)$$
$$sin 3x^2 = 3x^2-9/2 ; x^6+O(x^8)$$
then quotient is
$$frac-1/2 + 9/23-1 + O(x^2)$$
and therefore the limit is 2.
answered Jul 30 at 20:30
gammatester
15.7k21429
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
add a comment |Â
up vote
2
down vote
Use the Maclaurin series
$$cos x = 1-1/2 ; x^2+1/24 ; x^4 +O(x^6)$$
$$cos 3x = 1-9/2 ; x^2+27/8 ; x^4 +O(x^6)$$
$$sin x^2 = x^2-1/6; x^6+O(x^8)$$
$$sin 3x^2 = 3x^2-9/2 ; x^6+O(x^8)$$
then quotient is
$$frac-1/2 + 9/23-1 + O(x^2)$$
and therefore the limit is 2.
answered Jul 30 at 20:30
gammatester
15.7k21429
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Use the Maclaurin series
$$cos x = 1-1/2 ; x^2+1/24 ; x^4 +O(x^6)$$
$$cos 3x = 1-9/2 ; x^2+27/8 ; x^4 +O(x^6)$$
$$sin x^2 = x^2-1/6; x^6+O(x^8)$$
$$sin 3x^2 = 3x^2-9/2 ; x^6+O(x^8)$$
then quotient is
$$frac-1/2 + 9/23-1 + O(x^2)$$
and therefore the limit is 2.
answered Jul 30 at 20:30
gammatester
15.7k21429
Use the Maclaurin series
$$cos x = 1-1/2 ; x^2+1/24 ; x^4 +O(x^6)$$
$$cos 3x = 1-9/2 ; x^2+27/8 ; x^4 +O(x^6)$$
$$sin x^2 = x^2-1/6; x^6+O(x^8)$$
$$sin 3x^2 = 3x^2-9/2 ; x^6+O(x^8)$$
then quotient is
$$frac-1/2 + 9/23-1 + O(x^2)$$
and therefore the limit is 2.
answered Jul 30 at 20:30
gammatester
15.7k21429
answered Jul 30 at 20:30
gammatester
15.7k21429
answered Jul 30 at 20:30
gammatester
15.7k21429
answered Jul 30 at 20:30
gammatester
15.7k21429
15.7k21429
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
add a comment |Â
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
To simplify, It suffices to expand to the second order terms $x^2$.
– gimusi
Jul 30 at 20:38
add a comment |Â
up vote
2
down vote
Using l'Hôpital directly is very awkward.
The denominator has $x^2$, but it is immediate that
$$
lim_tto0fracsin3t-sin tt=3-1=2
$$
Consequently, also
$$
lim_xto0fracsin(3x^2)-sin(x^2)2x^2=1
$$
Good! Now we can write our limit in the form
$$
lim_xto0fraccos x-cos 3x2x^2frac2x^2sin(3x^2)-sin(x^2)
$$
The first fraction can be easily dealt with:
$$
lim_xto0fraccos x-cos 3x2x^2=
lim_xto0frac3sin3x-sin x4x=
lim_xto0frac9cos3x-cos x4=2
$$
For completeness, a much simpler strategy is using Taylor expansion:
$$
fraccos x-cos3xsin(3x^2)-sin(x^2)=
frac1-x^2/2-1+(3x)^2/2+o(x^2)3x^2-x^2+o(x^2)=
frac4+o(1)2+o(1)
$$
so the limit is $2$.
answered Jul 30 at 21:35
egreg
164k1180187
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
1
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
add a comment |Â
up vote
2
down vote
Using l'Hôpital directly is very awkward.
The denominator has $x^2$, but it is immediate that
$$
lim_tto0fracsin3t-sin tt=3-1=2
$$
Consequently, also
$$
lim_xto0fracsin(3x^2)-sin(x^2)2x^2=1
$$
Good! Now we can write our limit in the form
$$
lim_xto0fraccos x-cos 3x2x^2frac2x^2sin(3x^2)-sin(x^2)
$$
The first fraction can be easily dealt with:
$$
lim_xto0fraccos x-cos 3x2x^2=
lim_xto0frac3sin3x-sin x4x=
lim_xto0frac9cos3x-cos x4=2
$$
For completeness, a much simpler strategy is using Taylor expansion:
$$
fraccos x-cos3xsin(3x^2)-sin(x^2)=
frac1-x^2/2-1+(3x)^2/2+o(x^2)3x^2-x^2+o(x^2)=
frac4+o(1)2+o(1)
$$
so the limit is $2$.
answered Jul 30 at 21:35
egreg
164k1180187
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
1
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using l'Hôpital directly is very awkward.
The denominator has $x^2$, but it is immediate that
$$
lim_tto0fracsin3t-sin tt=3-1=2
$$
Consequently, also
$$
lim_xto0fracsin(3x^2)-sin(x^2)2x^2=1
$$
Good! Now we can write our limit in the form
$$
lim_xto0fraccos x-cos 3x2x^2frac2x^2sin(3x^2)-sin(x^2)
$$
The first fraction can be easily dealt with:
$$
lim_xto0fraccos x-cos 3x2x^2=
lim_xto0frac3sin3x-sin x4x=
lim_xto0frac9cos3x-cos x4=2
$$
For completeness, a much simpler strategy is using Taylor expansion:
$$
fraccos x-cos3xsin(3x^2)-sin(x^2)=
frac1-x^2/2-1+(3x)^2/2+o(x^2)3x^2-x^2+o(x^2)=
frac4+o(1)2+o(1)
$$
so the limit is $2$.
answered Jul 30 at 21:35
egreg
164k1180187
Using l'Hôpital directly is very awkward.
The denominator has $x^2$, but it is immediate that
$$
lim_tto0fracsin3t-sin tt=3-1=2
$$
Consequently, also
$$
lim_xto0fracsin(3x^2)-sin(x^2)2x^2=1
$$
Good! Now we can write our limit in the form
$$
lim_xto0fraccos x-cos 3x2x^2frac2x^2sin(3x^2)-sin(x^2)
$$
The first fraction can be easily dealt with:
$$
lim_xto0fraccos x-cos 3x2x^2=
lim_xto0frac3sin3x-sin x4x=
lim_xto0frac9cos3x-cos x4=2
$$
For completeness, a much simpler strategy is using Taylor expansion:
$$
fraccos x-cos3xsin(3x^2)-sin(x^2)=
frac1-x^2/2-1+(3x)^2/2+o(x^2)3x^2-x^2+o(x^2)=
frac4+o(1)2+o(1)
$$
so the limit is $2$.
answered Jul 30 at 21:35
egreg
164k1180187
answered Jul 30 at 21:35
egreg
164k1180187
answered Jul 30 at 21:35
egreg
164k1180187
answered Jul 30 at 21:35
egreg
164k1180187
164k1180187
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
1
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
add a comment |Â
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
1
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
Is this correct $o(1)→0$?
– Takahiro Waki
Jul 31 at 0:55
1
1
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
@TakahiroWaki Yes of course, recall that by definition $o(1)=1cdot omega(x)$ with $omega(x)to 0$.
– gimusi
Jul 31 at 5:39
add a comment |Â
up vote
0
down vote
$$
cos x-cos(3x) = cos xleft(sin^2x+3sin^2 xright)
$$
then
$$
fraccos x-cos (3x)sin (3x^2)-sin (x^2) = fracx^2x^2cos xleft(fracsin^2x+3sin^2 xsin (3x^2)-sin (x^2)right)=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right)
$$
hence
$$
lim_xto 0=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right) = 1cdotleft(frac1+33-1right) = 2
$$
answered Jul 30 at 22:01
Cesareo
5,5912412
add a comment |Â
up vote
0
down vote
$$
cos x-cos(3x) = cos xleft(sin^2x+3sin^2 xright)
$$
then
$$
fraccos x-cos (3x)sin (3x^2)-sin (x^2) = fracx^2x^2cos xleft(fracsin^2x+3sin^2 xsin (3x^2)-sin (x^2)right)=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right)
$$
hence
$$
lim_xto 0=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right) = 1cdotleft(frac1+33-1right) = 2
$$
answered Jul 30 at 22:01
Cesareo
5,5912412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$
cos x-cos(3x) = cos xleft(sin^2x+3sin^2 xright)
$$
then
$$
fraccos x-cos (3x)sin (3x^2)-sin (x^2) = fracx^2x^2cos xleft(fracsin^2x+3sin^2 xsin (3x^2)-sin (x^2)right)=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right)
$$
hence
$$
lim_xto 0=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right) = 1cdotleft(frac1+33-1right) = 2
$$
answered Jul 30 at 22:01
Cesareo
5,5912412
$$
cos x-cos(3x) = cos xleft(sin^2x+3sin^2 xright)
$$
then
$$
fraccos x-cos (3x)sin (3x^2)-sin (x^2) = fracx^2x^2cos xleft(fracsin^2x+3sin^2 xsin (3x^2)-sin (x^2)right)=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right)
$$
hence
$$
lim_xto 0=cos xleft(fracleft(fracsin xxright)^2+3left(fracsin xxright)^2frac3sin(3x^2)3x^2-fracsin (x^2)x^2right) = 1cdotleft(frac1+33-1right) = 2
$$
answered Jul 30 at 22:01
Cesareo
5,5912412
answered Jul 30 at 22:01
Cesareo
5,5912412
answered Jul 30 at 22:01
Cesareo
5,5912412
answered Jul 30 at 22:01
Cesareo
5,5912412
5,5912412
add a comment |Â
add a comment |Â
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