Mixing
$limlimits_{xto0}sin1/x$
$begingroup$
What is
$$limlimits_{xrightarrow0}{left( sin{frac{1}{x}}right)} $$?
Wolfram says "-1 to 1", but I don't know what that means.
In fact, I thought this limit didn't exist, so what does "-1 to 1" mean in this context?
real-analysis limits
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
$endgroup$
add a comment |
$begingroup$
What is
$$limlimits_{xrightarrow0}{left( sin{frac{1}{x}}right)} $$?
Wolfram says "-1 to 1", but I don't know what that means.
In fact, I thought this limit didn't exist, so what does "-1 to 1" mean in this context?
real-analysis limits
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
$endgroup$
1
$begingroup$
This limit does not exist. Wolfram Alpha says so too in parentheses.
$endgroup$
– bames
Feb 11 '18 at 6:36
2
$begingroup$
Thanks for all the answers. I understand what's going on now.
$endgroup$
– Stephen
Feb 11 '18 at 6:48
add a comment |
$begingroup$
What is
$$limlimits_{xrightarrow0}{left( sin{frac{1}{x}}right)} $$?
Wolfram says "-1 to 1", but I don't know what that means.
In fact, I thought this limit didn't exist, so what does "-1 to 1" mean in this context?
real-analysis limits
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
$endgroup$
What is
$$limlimits_{xrightarrow0}{left( sin{frac{1}{x}}right)} $$?
Wolfram says "-1 to 1", but I don't know what that means.
In fact, I thought this limit didn't exist, so what does "-1 to 1" mean in this context?
real-analysis limits
real-analysis limits
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
edited Jan 27 at 21:19
Stephen
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
asked Feb 11 '18 at 6:35
StephenStephen
1,3261819
1,3261819
1
$begingroup$
This limit does not exist. Wolfram Alpha says so too in parentheses.
$endgroup$
– bames
Feb 11 '18 at 6:36
2
$begingroup$
Thanks for all the answers. I understand what's going on now.
$endgroup$
– Stephen
Feb 11 '18 at 6:48
add a comment |
1
$begingroup$
This limit does not exist. Wolfram Alpha says so too in parentheses.
$endgroup$
– bames
Feb 11 '18 at 6:36
2
$begingroup$
Thanks for all the answers. I understand what's going on now.
$endgroup$
– Stephen
Feb 11 '18 at 6:48
1
1
$begingroup$
This limit does not exist. Wolfram Alpha says so too in parentheses.
$endgroup$
– bames
Feb 11 '18 at 6:36
$begingroup$
This limit does not exist. Wolfram Alpha says so too in parentheses.
$endgroup$
– bames
Feb 11 '18 at 6:36
2
2
$begingroup$
Thanks for all the answers. I understand what's going on now.
$endgroup$
– Stephen
Feb 11 '18 at 6:48
$begingroup$
Thanks for all the answers. I understand what's going on now.
$endgroup$
– Stephen
Feb 11 '18 at 6:48
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
For every $ain[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}rightarrow 0$ and $sin(1/x_{n})rightarrow a$.
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
$endgroup$
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
3
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
add a comment |
$begingroup$
$$Box nexists lim_{xto 0}bigg(sinfrac1xbigg).$$ Proof: Let $u = dfrac{1}{x}$ then $$lim_{xto 0}frac{1}{x} = infty$$ since $$lim_{xtoinfty}frac{1}{x} = 0.$$ Therefore, we get $$lim_{xto 0}bigg(sinfrac 1xbigg) = lim_{utoinfty}(sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $qquad qquadqquadqquadquad,,,,$
edited Jan 27 at 6:25
abc...
3,237738
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
$endgroup$
add a comment |
$begingroup$
The limit $$lim_{xrightarrow0}{left( sin{frac{1}{x}}right)}$$
does not exist.
Note that as x approaches $0$, $sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.
For example at $ x= frac {2}{(2n+1)pi } $ we have $sin(1/x)=pm 1.$
Thus there is no limit for $sin(1/x)$ as $x$ approaches $0$.
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
When $frac1xtoinfty$ it can be any value resulting from -1 to 1 because $-1<sin($any real number$)<1$ and the exact angle of $infty$ cannot be determined.
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
$endgroup$
add a comment |
$begingroup$
Take $a_n=dfrac{1}{npi}$. Clearly, $limlimits_{ntoinfty} a_n=0$ and $b_n=dfrac{1}{frac{1}{2}(4npi +pi)}$. Clearly, $limlimits_{ntoinfty}b_n=0$. Then, $limlimits_{ntoinfty}sinleft(dfrac{1}{a_n}right)=sin(npi)=0$ and $limlimits_{ntoinfty}sinleft(dfrac{1}{b_n}right)=sinleft(frac{1}{2}(4npi +pi)piright)=1$. Thus, the limit doesnt' exist.
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For every $ain[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}rightarrow 0$ and $sin(1/x_{n})rightarrow a$.
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
$endgroup$
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
3
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
add a comment |
$begingroup$
For every $ain[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}rightarrow 0$ and $sin(1/x_{n})rightarrow a$.
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
$endgroup$
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
3
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
add a comment |
$begingroup$
For every $ain[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}rightarrow 0$ and $sin(1/x_{n})rightarrow a$.
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
$endgroup$
For every $ain[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}rightarrow 0$ and $sin(1/x_{n})rightarrow a$.
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
answered Feb 11 '18 at 6:38
user284331user284331
35.4k31646
35.4k31646
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
3
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
add a comment |
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
3
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
$begingroup$
Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1.
$endgroup$
– Stephen
Feb 11 '18 at 6:40
3
3
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
Yes, the function assumes every point in $[-1,1]$ as adherent point.
$endgroup$
– user284331
Feb 11 '18 at 6:41
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
$begingroup$
In other words, $$-1leqslant sin x leqslant 1,$$ for all $xinmathbb{R}$.
$endgroup$
– user477343
Feb 20 '18 at 10:29
add a comment |
$begingroup$
$$Box nexists lim_{xto 0}bigg(sinfrac1xbigg).$$ Proof: Let $u = dfrac{1}{x}$ then $$lim_{xto 0}frac{1}{x} = infty$$ since $$lim_{xtoinfty}frac{1}{x} = 0.$$ Therefore, we get $$lim_{xto 0}bigg(sinfrac 1xbigg) = lim_{utoinfty}(sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $qquad qquadqquadqquadquad,,,,$
edited Jan 27 at 6:25
abc...
3,237738
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
$endgroup$
add a comment |
$begingroup$
$$Box nexists lim_{xto 0}bigg(sinfrac1xbigg).$$ Proof: Let $u = dfrac{1}{x}$ then $$lim_{xto 0}frac{1}{x} = infty$$ since $$lim_{xtoinfty}frac{1}{x} = 0.$$ Therefore, we get $$lim_{xto 0}bigg(sinfrac 1xbigg) = lim_{utoinfty}(sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $qquad qquadqquadqquadquad,,,,$
edited Jan 27 at 6:25
abc...
3,237738
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
$endgroup$
add a comment |
$begingroup$
$$Box nexists lim_{xto 0}bigg(sinfrac1xbigg).$$ Proof: Let $u = dfrac{1}{x}$ then $$lim_{xto 0}frac{1}{x} = infty$$ since $$lim_{xtoinfty}frac{1}{x} = 0.$$ Therefore, we get $$lim_{xto 0}bigg(sinfrac 1xbigg) = lim_{utoinfty}(sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $qquad qquadqquadqquadquad,,,,$
edited Jan 27 at 6:25
abc...
3,237738
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
$endgroup$
$$Box nexists lim_{xto 0}bigg(sinfrac1xbigg).$$ Proof: Let $u = dfrac{1}{x}$ then $$lim_{xto 0}frac{1}{x} = infty$$ since $$lim_{xtoinfty}frac{1}{x} = 0.$$ Therefore, we get $$lim_{xto 0}bigg(sinfrac 1xbigg) = lim_{utoinfty}(sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $qquad qquadqquadqquadquad,,,,$
edited Jan 27 at 6:25
abc...
3,237738
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
edited Jan 27 at 6:25
abc...
3,237738
edited Jan 27 at 6:25
abc...
3,237738
edited Jan 27 at 6:25
abc...
3,237738
3,237738
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
answered Feb 11 '18 at 6:46
user477343user477343
3,60831243
3,60831243
add a comment |
add a comment |
$begingroup$
The limit $$lim_{xrightarrow0}{left( sin{frac{1}{x}}right)}$$
does not exist.
Note that as x approaches $0$, $sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.
For example at $ x= frac {2}{(2n+1)pi } $ we have $sin(1/x)=pm 1.$
Thus there is no limit for $sin(1/x)$ as $x$ approaches $0$.
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
The limit $$lim_{xrightarrow0}{left( sin{frac{1}{x}}right)}$$
does not exist.
Note that as x approaches $0$, $sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.
For example at $ x= frac {2}{(2n+1)pi } $ we have $sin(1/x)=pm 1.$
Thus there is no limit for $sin(1/x)$ as $x$ approaches $0$.
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
The limit $$lim_{xrightarrow0}{left( sin{frac{1}{x}}right)}$$
does not exist.
Note that as x approaches $0$, $sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.
For example at $ x= frac {2}{(2n+1)pi } $ we have $sin(1/x)=pm 1.$
Thus there is no limit for $sin(1/x)$ as $x$ approaches $0$.
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
The limit $$lim_{xrightarrow0}{left( sin{frac{1}{x}}right)}$$
does not exist.
Note that as x approaches $0$, $sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.
For example at $ x= frac {2}{(2n+1)pi } $ we have $sin(1/x)=pm 1.$
Thus there is no limit for $sin(1/x)$ as $x$ approaches $0$.
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
edited Jan 27 at 7:51
Martin Sleziak
44.9k10122276
44.9k10122276
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Feb 11 '18 at 6:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
$begingroup$
When $frac1xtoinfty$ it can be any value resulting from -1 to 1 because $-1<sin($any real number$)<1$ and the exact angle of $infty$ cannot be determined.
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
$endgroup$
add a comment |
$begingroup$
When $frac1xtoinfty$ it can be any value resulting from -1 to 1 because $-1<sin($any real number$)<1$ and the exact angle of $infty$ cannot be determined.
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
$endgroup$
add a comment |
$begingroup$
When $frac1xtoinfty$ it can be any value resulting from -1 to 1 because $-1<sin($any real number$)<1$ and the exact angle of $infty$ cannot be determined.
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
$endgroup$
When $frac1xtoinfty$ it can be any value resulting from -1 to 1 because $-1<sin($any real number$)<1$ and the exact angle of $infty$ cannot be determined.
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
answered Feb 11 '18 at 6:42
Mehrdad ZandigoharMehrdad Zandigohar
1,528316
1,528316
add a comment |
add a comment |
$begingroup$
Take $a_n=dfrac{1}{npi}$. Clearly, $limlimits_{ntoinfty} a_n=0$ and $b_n=dfrac{1}{frac{1}{2}(4npi +pi)}$. Clearly, $limlimits_{ntoinfty}b_n=0$. Then, $limlimits_{ntoinfty}sinleft(dfrac{1}{a_n}right)=sin(npi)=0$ and $limlimits_{ntoinfty}sinleft(dfrac{1}{b_n}right)=sinleft(frac{1}{2}(4npi +pi)piright)=1$. Thus, the limit doesnt' exist.
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
$endgroup$
add a comment |
$begingroup$
Take $a_n=dfrac{1}{npi}$. Clearly, $limlimits_{ntoinfty} a_n=0$ and $b_n=dfrac{1}{frac{1}{2}(4npi +pi)}$. Clearly, $limlimits_{ntoinfty}b_n=0$. Then, $limlimits_{ntoinfty}sinleft(dfrac{1}{a_n}right)=sin(npi)=0$ and $limlimits_{ntoinfty}sinleft(dfrac{1}{b_n}right)=sinleft(frac{1}{2}(4npi +pi)piright)=1$. Thus, the limit doesnt' exist.
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
$endgroup$
add a comment |
$begingroup$
Take $a_n=dfrac{1}{npi}$. Clearly, $limlimits_{ntoinfty} a_n=0$ and $b_n=dfrac{1}{frac{1}{2}(4npi +pi)}$. Clearly, $limlimits_{ntoinfty}b_n=0$. Then, $limlimits_{ntoinfty}sinleft(dfrac{1}{a_n}right)=sin(npi)=0$ and $limlimits_{ntoinfty}sinleft(dfrac{1}{b_n}right)=sinleft(frac{1}{2}(4npi +pi)piright)=1$. Thus, the limit doesnt' exist.
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
$endgroup$
Take $a_n=dfrac{1}{npi}$. Clearly, $limlimits_{ntoinfty} a_n=0$ and $b_n=dfrac{1}{frac{1}{2}(4npi +pi)}$. Clearly, $limlimits_{ntoinfty}b_n=0$. Then, $limlimits_{ntoinfty}sinleft(dfrac{1}{a_n}right)=sin(npi)=0$ and $limlimits_{ntoinfty}sinleft(dfrac{1}{b_n}right)=sinleft(frac{1}{2}(4npi +pi)piright)=1$. Thus, the limit doesnt' exist.
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
answered Feb 11 '18 at 6:42
Carlos JiménezCarlos Jiménez
2,4121620
2,4121620
add a comment |
add a comment |
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1
$begingroup$
This limit does not exist. Wolfram Alpha says so too in parentheses.
$endgroup$
– bames
Feb 11 '18 at 6:36
2
$begingroup$
Thanks for all the answers. I understand what's going on now.
$endgroup$
– Stephen
Feb 11 '18 at 6:48