Squeeze theorem laws












2












$begingroup$


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







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  • 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
















2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 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














  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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 $lim_{xto 0}cos x=1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 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
      2 hours 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
      2 hours ago













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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    active

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    3












    $begingroup$

    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 $lim_{xto 0}cos x=1$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      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 $lim_{xto 0}cos x=1$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        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 $lim_{xto 0}cos x=1$.






        share|cite|improve this answer









        $endgroup$



        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 $lim_{xto 0}cos x=1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Paramanand SinghParamanand Singh

        50.2k556163




        50.2k556163























            2












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$









            • 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
              2 hours 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
              2 hours ago


















            2












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$









            • 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
              2 hours 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
              2 hours ago
















            2












            2








            2





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            LBJFSLBJFS

            1657




            1657








            • 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
              2 hours 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
              2 hours ago
















            • 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
              2 hours 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
              2 hours 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
            2 hours 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
            2 hours 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
            2 hours 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
            2 hours ago




















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