How does the squeeze theorem imply that $lim_{thetato 0}sintheta=0$ and $lim_{thetato 0}costheta=1$?












2












$begingroup$


I'm trying to understand how the sandwich/ squeeze theorem establishes that $lim_{thetato 0}sintheta = 0$ and $lim_{thetato 0}costheta = 1$.



Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.






calculus limits







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$endgroup$








  • 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
    21 hours ago
















2












$begingroup$


I'm trying to understand how the sandwich/ squeeze theorem establishes that $lim_{thetato 0}sintheta = 0$ and $lim_{thetato 0}costheta = 1$.



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
    21 hours ago














2












2








2





$begingroup$


I'm trying to understand how the sandwich/ squeeze theorem establishes that $lim_{thetato 0}sintheta = 0$ and $lim_{thetato 0}costheta = 1$.



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 $lim_{thetato 0}sintheta = 0$ and $lim_{thetato 0}costheta = 1$.



Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 18 hours ago









Asaf Karagila♦

305k33435765




305k33435765










asked 21 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
    21 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
    21 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
21 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
21 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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












    $begingroup$

    Edit: I corrected some inaccuracies pointed out in the comments.



    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 approch arbitrary close to the cell, then the prisoner must also approch arbitrary close to it (notice that the policemen don't need to go to the cell; this means the functions need not to be defined in the limit point). 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), $-lvert xrvertle sin xle lvert xrvert$ and this holds for any $x$ sufficiently small (actually for any $lvert xrvert $ in $]0,pi/2[$). 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.



    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











    $endgroup$









    • 3




      $begingroup$
      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
      20 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
      20 hours ago












    • $begingroup$
      @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
      $endgroup$
      – LBJFS
      16 hours ago










    • $begingroup$
      @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
      $endgroup$
      – LBJFS
      16 hours ago











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    2 Answers
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    $begingroup$

    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









    $endgroup$


















      4












      $begingroup$

      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









      $endgroup$
















        4












        4








        4





        $begingroup$

        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 21 hours ago









        Paramanand SinghParamanand Singh

        50.2k556163




        50.2k556163























            2












            $begingroup$

            Edit: I corrected some inaccuracies pointed out in the comments.



            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 approch arbitrary close to the cell, then the prisoner must also approch arbitrary close to it (notice that the policemen don't need to go to the cell; this means the functions need not to be defined in the limit point). 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), $-lvert xrvertle sin xle lvert xrvert$ and this holds for any $x$ sufficiently small (actually for any $lvert xrvert $ in $]0,pi/2[$). 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.



            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











            $endgroup$









            • 3




              $begingroup$
              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
              20 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
              20 hours ago












            • $begingroup$
              @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
              $endgroup$
              – LBJFS
              16 hours ago










            • $begingroup$
              @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
              $endgroup$
              – LBJFS
              16 hours ago
















            2












            $begingroup$

            Edit: I corrected some inaccuracies pointed out in the comments.



            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 approch arbitrary close to the cell, then the prisoner must also approch arbitrary close to it (notice that the policemen don't need to go to the cell; this means the functions need not to be defined in the limit point). 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), $-lvert xrvertle sin xle lvert xrvert$ and this holds for any $x$ sufficiently small (actually for any $lvert xrvert $ in $]0,pi/2[$). 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.



            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











            $endgroup$









            • 3




              $begingroup$
              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
              20 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
              20 hours ago












            • $begingroup$
              @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
              $endgroup$
              – LBJFS
              16 hours ago










            • $begingroup$
              @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
              $endgroup$
              – LBJFS
              16 hours ago














            2












            2








            2





            $begingroup$

            Edit: I corrected some inaccuracies pointed out in the comments.



            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 approch arbitrary close to the cell, then the prisoner must also approch arbitrary close to it (notice that the policemen don't need to go to the cell; this means the functions need not to be defined in the limit point). 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), $-lvert xrvertle sin xle lvert xrvert$ and this holds for any $x$ sufficiently small (actually for any $lvert xrvert $ in $]0,pi/2[$). 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.



            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











            $endgroup$



            Edit: I corrected some inaccuracies pointed out in the comments.



            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 approch arbitrary close to the cell, then the prisoner must also approch arbitrary close to it (notice that the policemen don't need to go to the cell; this means the functions need not to be defined in the limit point). 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), $-lvert xrvertle sin xle lvert xrvert$ and this holds for any $x$ sufficiently small (actually for any $lvert xrvert $ in $]0,pi/2[$). 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.



            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 14 hours ago

























            answered 21 hours ago









            LBJFSLBJFS

            2407




            2407








            • 3




              $begingroup$
              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
              20 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
              20 hours ago












            • $begingroup$
              @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
              $endgroup$
              – LBJFS
              16 hours ago










            • $begingroup$
              @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
              $endgroup$
              – LBJFS
              16 hours ago














            • 3




              $begingroup$
              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
              20 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
              20 hours ago












            • $begingroup$
              @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
              $endgroup$
              – LBJFS
              16 hours ago










            • $begingroup$
              @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
              $endgroup$
              – LBJFS
              16 hours ago








            3




            3




            $begingroup$
            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
            20 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
            20 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
            20 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
            20 hours ago














            $begingroup$
            @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
            $endgroup$
            – LBJFS
            16 hours ago




            $begingroup$
            @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-)
            $endgroup$
            – LBJFS
            16 hours ago












            $begingroup$
            @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
            $endgroup$
            – LBJFS
            16 hours ago




            $begingroup$
            @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out.
            $endgroup$
            – LBJFS
            16 hours ago


















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            Tabula Rosettana

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            Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a
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