Show that $g(x)=[frac1x]sin x$ has a limit in $x=0$.












2












$begingroup$


Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



( $[1/x]$ as greatest integer less than or equal to $1/x$)



I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



I know :
$$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



And I know :



$$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



I just don’t know how to use this stuffs to solve the question.



Thanks in advance fo the help .






calculus limits floor-function







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$endgroup$

















    2












    $begingroup$


    Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



    ( $[1/x]$ as greatest integer less than or equal to $1/x$)



    I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
    I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



    I know :
    $$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



    And I know :



    $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



    I just don’t know how to use this stuffs to solve the question.



    Thanks in advance fo the help .






    calculus limits floor-function







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



      ( $[1/x]$ as greatest integer less than or equal to $1/x$)



      I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
      I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



      I know :
      $$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



      And I know :



      $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



      I just don’t know how to use this stuffs to solve the question.



      Thanks in advance fo the help .






      calculus limits floor-function







      share|cite|improve this question











      $endgroup$




      Show that $g(x)=left[frac1x right]sin x$ has a limit in $x=0$.



      ( $[1/x]$ as greatest integer less than or equal to $1/x$)



      I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
      I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .



      I know :
      $$frac{1}{x}-1leleft[frac{1}{x}right]lefrac{1}{x},xnot=0.$$



      And I know :



      $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$



      I just don’t know how to use this stuffs to solve the question.



      Thanks in advance fo the help .






      calculus limits floor-function




      calculus limits floor-function






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 9 '18 at 3:37







      user98602

















      asked Dec 6 '18 at 18:18









      NegarNegar

      477




      477






















          3 Answers
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          active

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          2












          $begingroup$

          If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



          The function $sin$ is positive for small positive numbers and negative for small negative numbers.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



            Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              The OP is about the integer part [1/x].
              $endgroup$
              – gimusi
              Dec 6 '18 at 18:48










            • $begingroup$
              @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
              $endgroup$
              – J.G.
              Dec 6 '18 at 18:55



















            -3












            $begingroup$

            HINT



            You are mostly done, indeed since by floor function definition we have



            $$frac1x-1leleft[frac1xright]le frac1x$$



            then assuming wlog $0<x<1$



            $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



            and then use squeeze theorem.






            share|cite|improve this answer











            $endgroup$













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              3 Answers
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              2












              $begingroup$

              If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



              The function $sin$ is positive for small positive numbers and negative for small negative numbers.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



                The function $sin$ is positive for small positive numbers and negative for small negative numbers.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



                  The function $sin$ is positive for small positive numbers and negative for small negative numbers.






                  share|cite|improve this answer











                  $endgroup$



                  If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$left(frac{1}{x}-1right)sin(x)≤[frac{1}{x}]sin(x)≤frac{1}{x} sin(x)$$ since $$frac{1}{x}-1≤[frac{1}{x}]≤frac{1}{x},xnot=0.$$ Also for sufficiently small $x<0$ we have $$left(frac1x-1right)sin(x)≥[frac{1}{x}]sin(x)≥frac{1}{x} sin(x).$$ Now $$lim_{xrightarrow 0} frac{1}{x}sin(x)=1$$ and $$lim_{xrightarrow0}sin(x)=0.$$ Hence $$lim_{xrightarrow 0}[frac1x]sin(x)=1.$$



                  The function $sin$ is positive for small positive numbers and negative for small negative numbers.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 7 '18 at 8:40









                  Martin Sleziak

                  44.8k10119273




                  44.8k10119273










                  answered Dec 6 '18 at 18:36









                  UserSUserS

                  1,5541112




                  1,5541112























                      1












                      $begingroup$

                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55
















                      1












                      $begingroup$

                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55














                      1












                      1








                      1





                      $begingroup$

                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.






                      share|cite|improve this answer









                      $endgroup$



                      Let's start with the more famous problem, finding the limit of frac{sin x}{x}. Since $frac{sin x}{x}$ is even, we need only verify $lim_{xto 0^+}frac{sin x}{x}=1$. Draw a small angle $x$ between two length-$1$ radii $OA,,OB$ of a centre-$O$ circle $gamma$. Extend $OA$ through $A$ to meet the tangent-at-$B$ $ell$ to $gamma$ at $C$. The vertices $O,,A,,B$ are those of an area-$frac{1}{2}tan x$ triangle and an area-$frac{x}{2}$ sector, and the right-angled $triangle OAC$ has area $frac{1}{2}tan x$. Thus $$0lefrac{sin x}{x}le 1lefrac{tan x}{x}.$$Multiplying by $cos^{pm 1}x$ (which is positive), $$0lecos xlefrac{sin x}{x}le 1lefrac{tan x}{x}lesec x.$$(For example, get $cos xlefrac{sin x}{x}$ from $1lefrac{tan x}{x}$.) Since $cos xto 1$, the squeeze theorem finishes the proof.



                      Now for the problem at hand. We've subtract at most $|sin x|$ from the function considered above, but $sin xto 0$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 6 '18 at 18:36









                      J.G.J.G.

                      29.1k22845




                      29.1k22845








                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55














                      • 1




                        $begingroup$
                        The OP is about the integer part [1/x].
                        $endgroup$
                        – gimusi
                        Dec 6 '18 at 18:48










                      • $begingroup$
                        @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                        $endgroup$
                        – J.G.
                        Dec 6 '18 at 18:55








                      1




                      1




                      $begingroup$
                      The OP is about the integer part [1/x].
                      $endgroup$
                      – gimusi
                      Dec 6 '18 at 18:48




                      $begingroup$
                      The OP is about the integer part [1/x].
                      $endgroup$
                      – gimusi
                      Dec 6 '18 at 18:48












                      $begingroup$
                      @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                      $endgroup$
                      – J.G.
                      Dec 6 '18 at 18:55




                      $begingroup$
                      @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
                      $endgroup$
                      – J.G.
                      Dec 6 '18 at 18:55











                      -3












                      $begingroup$

                      HINT



                      You are mostly done, indeed since by floor function definition we have



                      $$frac1x-1leleft[frac1xright]le frac1x$$



                      then assuming wlog $0<x<1$



                      $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                      and then use squeeze theorem.






                      share|cite|improve this answer











                      $endgroup$


















                        -3












                        $begingroup$

                        HINT



                        You are mostly done, indeed since by floor function definition we have



                        $$frac1x-1leleft[frac1xright]le frac1x$$



                        then assuming wlog $0<x<1$



                        $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                        and then use squeeze theorem.






                        share|cite|improve this answer











                        $endgroup$
















                          -3












                          -3








                          -3





                          $begingroup$

                          HINT



                          You are mostly done, indeed since by floor function definition we have



                          $$frac1x-1leleft[frac1xright]le frac1x$$



                          then assuming wlog $0<x<1$



                          $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                          and then use squeeze theorem.






                          share|cite|improve this answer











                          $endgroup$



                          HINT



                          You are mostly done, indeed since by floor function definition we have



                          $$frac1x-1leleft[frac1xright]le frac1x$$



                          then assuming wlog $0<x<1$



                          $$ left(frac1x-1right)cdotsin xleleft[frac1xright]cdotsin xle frac1xcdotsin x$$



                          and then use squeeze theorem.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 7 '18 at 8:49

























                          answered Dec 6 '18 at 18:39









                          gimusigimusi

                          92.9k84494




                          92.9k84494






























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