Compute the limit:
$begingroup$
$$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
I tried:
$$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
$$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
Somewhere I over-complicated and I don't know how to continue or the other route that I should take...
limits trigonometry
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
$endgroup$
add a comment |
$begingroup$
$$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
I tried:
$$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
$$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
Somewhere I over-complicated and I don't know how to continue or the other route that I should take...
limits trigonometry
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
$endgroup$
add a comment |
$begingroup$
$$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
I tried:
$$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
$$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
Somewhere I over-complicated and I don't know how to continue or the other route that I should take...
limits trigonometry
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
$endgroup$
$$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
I tried:
$$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
$$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
Somewhere I over-complicated and I don't know how to continue or the other route that I should take...
limits trigonometry
limits trigonometry
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
edited Dec 9 '18 at 0:48
Key Flex
8,61861233
8,61861233
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
asked Dec 8 '18 at 16:27
Bili DebiliBili Debili
1428
1428
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
Again apply L'Hopital's Rule
$$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
$endgroup$
add a comment |
$begingroup$
Hint: Write
$$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
&= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
&= lim_{x to 0} frac{frac32x^2}{2x^2}\
&= frac34end{align}
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
1
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
add a comment |
$begingroup$
We have that
$${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$
then refer to standard limits
- $frac{1-cos(x)}{x^2}to frac12$
- $frac{2x}{sin(2x)}to 1$
and simply
- $1+cos x+cos^2(x)to 3$
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
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5 Answers
5
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
Again apply L'Hopital's Rule
$$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
Again apply L'Hopital's Rule
$$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
Again apply L'Hopital's Rule
$$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
$endgroup$
$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
Again apply L'Hopital's Rule
$$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
answered Dec 8 '18 at 16:41
Key FlexKey Flex
8,61861233
8,61861233
add a comment |
add a comment |
$begingroup$
Hint: Write
$$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
Hint: Write
$$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
Hint: Write
$$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
Hint: Write
$$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 8 '18 at 16:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
77.8k42866
add a comment |
add a comment |
$begingroup$
begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
&= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
&= lim_{x to 0} frac{frac32x^2}{2x^2}\
&= frac34end{align}
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
1
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
add a comment |
$begingroup$
begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
&= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
&= lim_{x to 0} frac{frac32x^2}{2x^2}\
&= frac34end{align}
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
1
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
add a comment |
$begingroup$
begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
&= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
&= lim_{x to 0} frac{frac32x^2}{2x^2}\
&= frac34end{align}
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
&= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
&= lim_{x to 0} frac{frac32x^2}{2x^2}\
&= frac34end{align}
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 8 '18 at 16:37
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
1
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
add a comment |
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
1
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
What is the step in between the first computation, where you get rid of trigonometric identities
$endgroup$
– Bili Debili
Dec 8 '18 at 16:44
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
$begingroup$
taylor series, just drop the higher order terms.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 16:47
1
1
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
$begingroup$
I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
$endgroup$
– gimusi
Dec 8 '18 at 17:00
add a comment |
$begingroup$
We have that
$${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$
then refer to standard limits
- $frac{1-cos(x)}{x^2}to frac12$
- $frac{2x}{sin(2x)}to 1$
and simply
- $1+cos x+cos^2(x)to 3$
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
We have that
$${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$
then refer to standard limits
- $frac{1-cos(x)}{x^2}to frac12$
- $frac{2x}{sin(2x)}to 1$
and simply
- $1+cos x+cos^2(x)to 3$
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
We have that
$${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$
then refer to standard limits
- $frac{1-cos(x)}{x^2}to frac12$
- $frac{2x}{sin(2x)}to 1$
and simply
- $1+cos x+cos^2(x)to 3$
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
$endgroup$
We have that
$${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$
then refer to standard limits
- $frac{1-cos(x)}{x^2}to frac12$
- $frac{2x}{sin(2x)}to 1$
and simply
- $1+cos x+cos^2(x)to 3$
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
edited Dec 8 '18 at 16:59
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
answered Dec 8 '18 at 16:30
gimusigimusi
93k84594
93k84594
add a comment |
add a comment |
$begingroup$
Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
$begingroup$
Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
$begingroup$
Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
answered Dec 9 '18 at 5:06
Paramanand SinghParamanand Singh
50.7k557168
50.7k557168
add a comment |
add a comment |
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