Does the limit of $sin (pi/x)$ converge or diverge?
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Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$
I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.
real-analysis sequences-and-series limits
edited Dec 2 at 21:33
Key Flex
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asked Dec 2 at 21:31
Ty Johnson
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add a comment |
up vote
0
down vote
favorite
Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$
I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.
real-analysis sequences-and-series limits
edited Dec 2 at 21:33
Key Flex
7,08441229
asked Dec 2 at 21:31
Ty Johnson
283
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
– Michael Lee
Dec 2 at 21:36
You can use an alternating sequence
– Fakemistake
Dec 2 at 21:37
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
– Paramanand Singh
Dec 3 at 2:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$
I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.
real-analysis sequences-and-series limits
edited Dec 2 at 21:33
Key Flex
7,08441229
asked Dec 2 at 21:31
Ty Johnson
283
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Does the limit converge or diverge? Justify your answer. $$lim_{xto0} [sin(pi/x)] $$
I know for sure that the limit diverges. I know I can use sub-sequences because it is a sine function, and I can show that it has two subsequential limits, thus the limit diverges. I just can't figure out what subsequences to use and how to show it.
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Dec 2 at 21:33
Key Flex
7,08441229
asked Dec 2 at 21:31
Ty Johnson
283
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 21:33
Key Flex
7,08441229
asked Dec 2 at 21:31
Ty Johnson
283
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 21:33
Key Flex
7,08441229
edited Dec 2 at 21:33
Key Flex
7,08441229
edited Dec 2 at 21:33
Key Flex
7,08441229
7,08441229
asked Dec 2 at 21:31
Ty Johnson
283
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 21:31
Ty Johnson
283
asked Dec 2 at 21:31
Ty Johnson
283
283
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
– Michael Lee
Dec 2 at 21:36
You can use an alternating sequence
– Fakemistake
Dec 2 at 21:37
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
– Paramanand Singh
Dec 3 at 2:04
add a comment |
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
– Michael Lee
Dec 2 at 21:36
You can use an alternating sequence
– Fakemistake
Dec 2 at 21:37
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
– Paramanand Singh
Dec 3 at 2:04
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
– Michael Lee
Dec 2 at 21:36
First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
– Michael Lee
Dec 2 at 21:36
You can use an alternating sequence
– Fakemistake
Dec 2 at 21:37
You can use an alternating sequence
– Fakemistake
Dec 2 at 21:37
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
– Paramanand Singh
Dec 3 at 2:04
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
– Paramanand Singh
Dec 3 at 2:04
add a comment |
4 Answers
4
active
oldest
votes
up vote
1
down vote
hint
$$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$
$$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
add a comment |
up vote
0
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Given $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$
Use the subsequence $x=dfrac{1}{2n}$
$$lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)=lim_{nrightarrow0}left[sinleft(dfrac{pi}{left(dfrac{1}{2n}right)}right)right].....$$
Therefore, $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$ diverges at $x=0$
answered Dec 2 at 21:38
Key Flex
7,08441229
add a comment |
up vote
0
down vote
Let consider the case $x=frac1yto 0^+ quad y to infty$ then
$$sin (pi/x)=sin (ypi )$$
and then the subsequence
$$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$
answered Dec 2 at 21:59
gimusi
90.7k74495
add a comment |
up vote
0
down vote
Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?
You should try to take a look on the graph of the function.
answered Dec 2 at 21:39
Célio Augusto
152
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
hint
$$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$
$$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
add a comment |
up vote
1
down vote
hint
$$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$
$$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
add a comment |
up vote
1
down vote
up vote
1
down vote
hint
$$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$
$$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
hint
$$sin(npi)=0=sin(frac{pi}{frac{1}{n}})$$
$$sin(fracpi 2+2npi)=1=sin(frac{pi}{...})$$
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
answered Dec 2 at 21:37
hamam_Abdallah
37.3k21634
37.3k21634
add a comment |
add a comment |
up vote
0
down vote
Given $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$
Use the subsequence $x=dfrac{1}{2n}$
$$lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)=lim_{nrightarrow0}left[sinleft(dfrac{pi}{left(dfrac{1}{2n}right)}right)right].....$$
Therefore, $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$ diverges at $x=0$
answered Dec 2 at 21:38
Key Flex
7,08441229
add a comment |
up vote
0
down vote
Given $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$
Use the subsequence $x=dfrac{1}{2n}$
$$lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)=lim_{nrightarrow0}left[sinleft(dfrac{pi}{left(dfrac{1}{2n}right)}right)right].....$$
Therefore, $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$ diverges at $x=0$
answered Dec 2 at 21:38
Key Flex
7,08441229
add a comment |
up vote
0
down vote
up vote
0
down vote
Given $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$
Use the subsequence $x=dfrac{1}{2n}$
$$lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)=lim_{nrightarrow0}left[sinleft(dfrac{pi}{left(dfrac{1}{2n}right)}right)right].....$$
Therefore, $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$ diverges at $x=0$
answered Dec 2 at 21:38
Key Flex
7,08441229
Given $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$
Use the subsequence $x=dfrac{1}{2n}$
$$lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)=lim_{nrightarrow0}left[sinleft(dfrac{pi}{left(dfrac{1}{2n}right)}right)right].....$$
Therefore, $lim_{xrightarrow0}sinleft(dfrac{pi}{x}right)$ diverges at $x=0$
answered Dec 2 at 21:38
Key Flex
7,08441229
answered Dec 2 at 21:38
Key Flex
7,08441229
answered Dec 2 at 21:38
Key Flex
7,08441229
answered Dec 2 at 21:38
Key Flex
7,08441229
7,08441229
add a comment |
add a comment |
up vote
0
down vote
Let consider the case $x=frac1yto 0^+ quad y to infty$ then
$$sin (pi/x)=sin (ypi )$$
and then the subsequence
$$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$
answered Dec 2 at 21:59
gimusi
90.7k74495
add a comment |
up vote
0
down vote
Let consider the case $x=frac1yto 0^+ quad y to infty$ then
$$sin (pi/x)=sin (ypi )$$
and then the subsequence
$$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$
answered Dec 2 at 21:59
gimusi
90.7k74495
add a comment |
up vote
0
down vote
up vote
0
down vote
Let consider the case $x=frac1yto 0^+ quad y to infty$ then
$$sin (pi/x)=sin (ypi )$$
and then the subsequence
$$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$
answered Dec 2 at 21:59
gimusi
90.7k74495
Let consider the case $x=frac1yto 0^+ quad y to infty$ then
$$sin (pi/x)=sin (ypi )$$
and then the subsequence
$$y_n=frac12+n implies sin (pi y_n)=cos(npi) to pm 1$$
answered Dec 2 at 21:59
gimusi
90.7k74495
answered Dec 2 at 21:59
gimusi
90.7k74495
answered Dec 2 at 21:59
gimusi
90.7k74495
answered Dec 2 at 21:59
gimusi
90.7k74495
90.7k74495
add a comment |
add a comment |
up vote
0
down vote
Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?
You should try to take a look on the graph of the function.
answered Dec 2 at 21:39
Célio Augusto
152
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
add a comment |
up vote
0
down vote
Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?
You should try to take a look on the graph of the function.
answered Dec 2 at 21:39
Célio Augusto
152
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
add a comment |
up vote
0
down vote
up vote
0
down vote
Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?
You should try to take a look on the graph of the function.
answered Dec 2 at 21:39
Célio Augusto
152
Since the sine is a periodic function, a good approach is trying sequences $y_n$ and $x_n$ satisfying both $f(x_n)$ and $f(y_)n$ are constants, but $x_n neq y_n$. Can you guess such sequences?
You should try to take a look on the graph of the function.
answered Dec 2 at 21:39
Célio Augusto
152
edited Dec 4 at 22:19
answered Dec 2 at 21:39
Célio Augusto
152
answered Dec 2 at 21:39
Célio Augusto
152
answered Dec 2 at 21:39
Célio Augusto
152
152
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
add a comment |
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
$x_n$ and $y_n$ are constants?
– Ty Johnson
Dec 2 at 22:25
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
I am sorry, the fuction applied at those points is constanr
– Célio Augusto
Dec 4 at 22:18
add a comment |
Ty Johnson is a new contributor. Be nice, and check out our Code of Conduct.
Ty Johnson is a new contributor. Be nice, and check out our Code of Conduct.
Ty Johnson is a new contributor. Be nice, and check out our Code of Conduct.
Ty Johnson is a new contributor. Be nice, and check out our Code of Conduct.
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First, what about $x_n = 1/n$? Then $sin(pi/x_n) = sin(npi) = 0$ for all $n$. Then, try $x_n = 2/(4n-3)$, which gives $sin(pi/x_n) = sin((4n-3)pi/2) = 1$ for all $n$.
– Michael Lee
Dec 2 at 21:36
You can use an alternating sequence
– Fakemistake
Dec 2 at 21:37
It appears that by "diverge" you mean "not converge" and then in this sense the function does diverge. Some books use "diverge" to mean "diverge to $pminfty" $ and there is another term "oscillate" to describe the behavior of the function in this question. I prefer to say $sin(pi/x) $ oscillates finitely as $xto 0$.
– Paramanand Singh
Dec 3 at 2:04