Why does $limlimits_{thetatoinfty}nsinfrac {theta}n=theta$












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Questions:





  1. Why do we have$$limlimits_{thetatoinfty}nsindfrac {theta}n=thetatag1$$

  2. How do we prove $(1)$?


I started off with the well known limit:$$limlimits_{thetato 0}dfrac {sintheta}{theta}=1tag2$$
And substituted $theta:=dfrac theta n$ into $(1)$ to get$$limlimits_{frac theta nto 0}dfrac {sindfrac theta n}{dfrac theta n}=limlimits_{frac theta nto0}dfrac {nsinfrac theta n}{theta}tag3$$
However after that, I'm not sure what to do. Apparently, the RHS of $(1)$ has a limit of $1$, so that the limit of $nsinfrac theta n=theta$.






limits







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  • 2




    $begingroup$
    Do you mean $lim_{ntoinfty}$?
    $endgroup$
    – carmichael561
    Feb 15 '17 at 1:14






  • 2




    $begingroup$
    It is false that $lim_{thetatoinfty}dots=theta$, since $theta$ itself goes to infinity, but $sin$ is bounded. Perhaps you meant the limit as $ntoinfty$?
    $endgroup$
    – Simply Beautiful Art
    Feb 15 '17 at 1:15










  • $begingroup$
    The expression $lim_{thetatodots}dots = theta$ isn't even a valid equation because $theta$ isn't a free variable of the left hand side.
    $endgroup$
    – mniip
    Feb 15 '17 at 1:26


















0












$begingroup$



Questions:





  1. Why do we have$$limlimits_{thetatoinfty}nsindfrac {theta}n=thetatag1$$

  2. How do we prove $(1)$?


I started off with the well known limit:$$limlimits_{thetato 0}dfrac {sintheta}{theta}=1tag2$$
And substituted $theta:=dfrac theta n$ into $(1)$ to get$$limlimits_{frac theta nto 0}dfrac {sindfrac theta n}{dfrac theta n}=limlimits_{frac theta nto0}dfrac {nsinfrac theta n}{theta}tag3$$
However after that, I'm not sure what to do. Apparently, the RHS of $(1)$ has a limit of $1$, so that the limit of $nsinfrac theta n=theta$.






limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Do you mean $lim_{ntoinfty}$?
    $endgroup$
    – carmichael561
    Feb 15 '17 at 1:14






  • 2




    $begingroup$
    It is false that $lim_{thetatoinfty}dots=theta$, since $theta$ itself goes to infinity, but $sin$ is bounded. Perhaps you meant the limit as $ntoinfty$?
    $endgroup$
    – Simply Beautiful Art
    Feb 15 '17 at 1:15










  • $begingroup$
    The expression $lim_{thetatodots}dots = theta$ isn't even a valid equation because $theta$ isn't a free variable of the left hand side.
    $endgroup$
    – mniip
    Feb 15 '17 at 1:26
















0












0








0


0



$begingroup$



Questions:





  1. Why do we have$$limlimits_{thetatoinfty}nsindfrac {theta}n=thetatag1$$

  2. How do we prove $(1)$?


I started off with the well known limit:$$limlimits_{thetato 0}dfrac {sintheta}{theta}=1tag2$$
And substituted $theta:=dfrac theta n$ into $(1)$ to get$$limlimits_{frac theta nto 0}dfrac {sindfrac theta n}{dfrac theta n}=limlimits_{frac theta nto0}dfrac {nsinfrac theta n}{theta}tag3$$
However after that, I'm not sure what to do. Apparently, the RHS of $(1)$ has a limit of $1$, so that the limit of $nsinfrac theta n=theta$.






limits







share|cite|improve this question











$endgroup$





Questions:





  1. Why do we have$$limlimits_{thetatoinfty}nsindfrac {theta}n=thetatag1$$

  2. How do we prove $(1)$?


I started off with the well known limit:$$limlimits_{thetato 0}dfrac {sintheta}{theta}=1tag2$$
And substituted $theta:=dfrac theta n$ into $(1)$ to get$$limlimits_{frac theta nto 0}dfrac {sindfrac theta n}{dfrac theta n}=limlimits_{frac theta nto0}dfrac {nsinfrac theta n}{theta}tag3$$
However after that, I'm not sure what to do. Apparently, the RHS of $(1)$ has a limit of $1$, so that the limit of $nsinfrac theta n=theta$.






limits




limits






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edited Dec 13 '18 at 14:38









Martin Sleziak

44.8k10118272




44.8k10118272










asked Feb 15 '17 at 1:13









FrankFrank

3,7751632




3,7751632








  • 2




    $begingroup$
    Do you mean $lim_{ntoinfty}$?
    $endgroup$
    – carmichael561
    Feb 15 '17 at 1:14






  • 2




    $begingroup$
    It is false that $lim_{thetatoinfty}dots=theta$, since $theta$ itself goes to infinity, but $sin$ is bounded. Perhaps you meant the limit as $ntoinfty$?
    $endgroup$
    – Simply Beautiful Art
    Feb 15 '17 at 1:15










  • $begingroup$
    The expression $lim_{thetatodots}dots = theta$ isn't even a valid equation because $theta$ isn't a free variable of the left hand side.
    $endgroup$
    – mniip
    Feb 15 '17 at 1:26
















  • 2




    $begingroup$
    Do you mean $lim_{ntoinfty}$?
    $endgroup$
    – carmichael561
    Feb 15 '17 at 1:14






  • 2




    $begingroup$
    It is false that $lim_{thetatoinfty}dots=theta$, since $theta$ itself goes to infinity, but $sin$ is bounded. Perhaps you meant the limit as $ntoinfty$?
    $endgroup$
    – Simply Beautiful Art
    Feb 15 '17 at 1:15










  • $begingroup$
    The expression $lim_{thetatodots}dots = theta$ isn't even a valid equation because $theta$ isn't a free variable of the left hand side.
    $endgroup$
    – mniip
    Feb 15 '17 at 1:26










2




2




$begingroup$
Do you mean $lim_{ntoinfty}$?
$endgroup$
– carmichael561
Feb 15 '17 at 1:14




$begingroup$
Do you mean $lim_{ntoinfty}$?
$endgroup$
– carmichael561
Feb 15 '17 at 1:14




2




2




$begingroup$
It is false that $lim_{thetatoinfty}dots=theta$, since $theta$ itself goes to infinity, but $sin$ is bounded. Perhaps you meant the limit as $ntoinfty$?
$endgroup$
– Simply Beautiful Art
Feb 15 '17 at 1:15




$begingroup$
It is false that $lim_{thetatoinfty}dots=theta$, since $theta$ itself goes to infinity, but $sin$ is bounded. Perhaps you meant the limit as $ntoinfty$?
$endgroup$
– Simply Beautiful Art
Feb 15 '17 at 1:15












$begingroup$
The expression $lim_{thetatodots}dots = theta$ isn't even a valid equation because $theta$ isn't a free variable of the left hand side.
$endgroup$
– mniip
Feb 15 '17 at 1:26






$begingroup$
The expression $lim_{thetatodots}dots = theta$ isn't even a valid equation because $theta$ isn't a free variable of the left hand side.
$endgroup$
– mniip
Feb 15 '17 at 1:26












3 Answers
3






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$begingroup$

I think you meant $lim_{nto infty}n sinleft(frac{theta}{n}right) = theta$, which you basically proved is true:



$lim_{nto infty}n sinleft(frac{theta}{n}right) = lim_{nto infty} left( thetacdot frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot lim_{nto infty} left( frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot 1 =theta$






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$endgroup$





















    1












    $begingroup$

    Assuming you meant $lim_{ntoinfty}$, note that by letting $u=theta/n$, we get



    $$lim_{ntoinfty}nsinfractheta n=thetalim_{uto0^{pm}}frac{sin u}u=thetatimes1$$



    where the side we approach depends on the sign of theta, but it doesn't affect the limit.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Or, with equivalents, $sindfractheta nsim_infty dfractheta n$, so
      $$nsinfractheta nsim_infty nfractheta n=theta.$$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

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        2












        $begingroup$

        I think you meant $lim_{nto infty}n sinleft(frac{theta}{n}right) = theta$, which you basically proved is true:



        $lim_{nto infty}n sinleft(frac{theta}{n}right) = lim_{nto infty} left( thetacdot frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot lim_{nto infty} left( frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot 1 =theta$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          I think you meant $lim_{nto infty}n sinleft(frac{theta}{n}right) = theta$, which you basically proved is true:



          $lim_{nto infty}n sinleft(frac{theta}{n}right) = lim_{nto infty} left( thetacdot frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot lim_{nto infty} left( frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot 1 =theta$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            I think you meant $lim_{nto infty}n sinleft(frac{theta}{n}right) = theta$, which you basically proved is true:



            $lim_{nto infty}n sinleft(frac{theta}{n}right) = lim_{nto infty} left( thetacdot frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot lim_{nto infty} left( frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot 1 =theta$






            share|cite|improve this answer









            $endgroup$



            I think you meant $lim_{nto infty}n sinleft(frac{theta}{n}right) = theta$, which you basically proved is true:



            $lim_{nto infty}n sinleft(frac{theta}{n}right) = lim_{nto infty} left( thetacdot frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot lim_{nto infty} left( frac{sinleft(frac{theta}{n}right)}{frac{theta}{n}} right) = theta cdot 1 =theta$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 15 '17 at 1:27









            joebjoeb

            2,121211




            2,121211























                1












                $begingroup$

                Assuming you meant $lim_{ntoinfty}$, note that by letting $u=theta/n$, we get



                $$lim_{ntoinfty}nsinfractheta n=thetalim_{uto0^{pm}}frac{sin u}u=thetatimes1$$



                where the side we approach depends on the sign of theta, but it doesn't affect the limit.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Assuming you meant $lim_{ntoinfty}$, note that by letting $u=theta/n$, we get



                  $$lim_{ntoinfty}nsinfractheta n=thetalim_{uto0^{pm}}frac{sin u}u=thetatimes1$$



                  where the side we approach depends on the sign of theta, but it doesn't affect the limit.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Assuming you meant $lim_{ntoinfty}$, note that by letting $u=theta/n$, we get



                    $$lim_{ntoinfty}nsinfractheta n=thetalim_{uto0^{pm}}frac{sin u}u=thetatimes1$$



                    where the side we approach depends on the sign of theta, but it doesn't affect the limit.






                    share|cite|improve this answer









                    $endgroup$



                    Assuming you meant $lim_{ntoinfty}$, note that by letting $u=theta/n$, we get



                    $$lim_{ntoinfty}nsinfractheta n=thetalim_{uto0^{pm}}frac{sin u}u=thetatimes1$$



                    where the side we approach depends on the sign of theta, but it doesn't affect the limit.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 15 '17 at 1:28









                    Simply Beautiful ArtSimply Beautiful Art

                    50.5k578182




                    50.5k578182























                        0












                        $begingroup$

                        Or, with equivalents, $sindfractheta nsim_infty dfractheta n$, so
                        $$nsinfractheta nsim_infty nfractheta n=theta.$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Or, with equivalents, $sindfractheta nsim_infty dfractheta n$, so
                          $$nsinfractheta nsim_infty nfractheta n=theta.$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Or, with equivalents, $sindfractheta nsim_infty dfractheta n$, so
                            $$nsinfractheta nsim_infty nfractheta n=theta.$$






                            share|cite|improve this answer









                            $endgroup$



                            Or, with equivalents, $sindfractheta nsim_infty dfractheta n$, so
                            $$nsinfractheta nsim_infty nfractheta n=theta.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 15 '17 at 1:56









                            BernardBernard

                            121k740116




                            121k740116






























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