Squeeze theorem laws
$begingroup$
I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.
Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.
calculus limits
asked 2 hours ago
JamesJames
163
$endgroup$
add a comment |
$begingroup$
I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.
Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.
calculus limits
asked 2 hours ago
JamesJames
163
$endgroup$
1
$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago
add a comment |
$begingroup$
I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.
Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.
calculus limits
asked 2 hours ago
JamesJames
163
$endgroup$
I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.
Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.
calculus limits
calculus limits
asked 2 hours ago
JamesJames
163
asked 2 hours ago
JamesJames
163
asked 2 hours ago
JamesJames
163
asked 2 hours ago
JamesJames
163
asked 2 hours ago
JamesJames
163
163
1
$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago
add a comment |
1
$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago
1
1
$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago
$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.
We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.
For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
$endgroup$
add a comment |
$begingroup$
I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.
Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.
Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.
answered 2 hours ago
LBJFSLBJFS
1557
$endgroup$
1
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
1
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.
We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.
For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
$endgroup$
add a comment |
$begingroup$
The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.
We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.
For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
$endgroup$
add a comment |
$begingroup$
The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.
We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.
For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
$endgroup$
The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.
We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.
For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
answered 2 hours ago
Paramanand SinghParamanand Singh
50.2k556163
50.2k556163
add a comment |
add a comment |
$begingroup$
I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.
Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.
Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.
answered 2 hours ago
LBJFSLBJFS
1557
$endgroup$
1
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
1
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
$begingroup$
I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.
Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.
Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.
answered 2 hours ago
LBJFSLBJFS
1557
$endgroup$
1
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
1
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
$begingroup$
I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.
Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.
Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.
answered 2 hours ago
LBJFSLBJFS
1557
$endgroup$
I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.
Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.
Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.
answered 2 hours ago
LBJFSLBJFS
1557
edited 1 hour ago
answered 2 hours ago
LBJFSLBJFS
1557
answered 2 hours ago
LBJFSLBJFS
1557
answered 2 hours ago
LBJFSLBJFS
1557
1557
1
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
1
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
1
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
1
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
1
1
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
$begingroup$
Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
$endgroup$
– Brian Tung
1 hour ago
1
1
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
$begingroup$
There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
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$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago