How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$












1












$begingroup$


How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$ without lhopital rule?

I know when I use lhopital I easy get
$$ lim_{hrightarrow 0}frac{cos(ah)a}{1} = a$$ but I don't know how to behave without that way






real-analysis limits limits-without-lhopital







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    1












    $begingroup$


    How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$ without lhopital rule?

    I know when I use lhopital I easy get
    $$ lim_{hrightarrow 0}frac{cos(ah)a}{1} = a$$ but I don't know how to behave without that way






    real-analysis limits limits-without-lhopital







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$ without lhopital rule?

      I know when I use lhopital I easy get
      $$ lim_{hrightarrow 0}frac{cos(ah)a}{1} = a$$ but I don't know how to behave without that way






      real-analysis limits limits-without-lhopital







      share|cite|improve this question









      $endgroup$




      How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$ without lhopital rule?

      I know when I use lhopital I easy get
      $$ lim_{hrightarrow 0}frac{cos(ah)a}{1} = a$$ but I don't know how to behave without that way






      real-analysis limits limits-without-lhopital




      real-analysis limits limits-without-lhopital






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 28 at 12:18









      VirtualUserVirtualUser

      1,120117




      1,120117






















          2 Answers
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          4












          $begingroup$

          Hint:



          $$frac{sin(ha)}{h} = acdotfrac{sin(ha)}{ha}$$



          Also, remember what $$lim_{xto 0}frac{sin(x)}{x}$$ is equal to?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for show that simple trick!
            $endgroup$
            – VirtualUser
            Jan 28 at 12:45



















          0












          $begingroup$

          Notice that $sin(x) = sum_{k = 0}^infty (-1)^k frac{x^{2k + 1}}{(2k + 1)!}$ for all $x in mathbb R$. Thus,
          $$ frac{sin(ah)}{h} = frac 1 h sum_{k = 0}^infty (-1)^k frac{(ah)^{2k + 1}}{(2k + 1)!} = sum_{k = 0}^infty (-1)^k frac{a^{2k + 1} h^{2k}}{(2k + 1)!} longrightarrow a$$
          as $h to 0$ by dominated convergence for example.






          share|cite|improve this answer









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            2 Answers
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            active

            oldest

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            2 Answers
            2






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            active

            oldest

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            4












            $begingroup$

            Hint:



            $$frac{sin(ha)}{h} = acdotfrac{sin(ha)}{ha}$$



            Also, remember what $$lim_{xto 0}frac{sin(x)}{x}$$ is equal to?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for show that simple trick!
              $endgroup$
              – VirtualUser
              Jan 28 at 12:45
















            4












            $begingroup$

            Hint:



            $$frac{sin(ha)}{h} = acdotfrac{sin(ha)}{ha}$$



            Also, remember what $$lim_{xto 0}frac{sin(x)}{x}$$ is equal to?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for show that simple trick!
              $endgroup$
              – VirtualUser
              Jan 28 at 12:45














            4












            4








            4





            $begingroup$

            Hint:



            $$frac{sin(ha)}{h} = acdotfrac{sin(ha)}{ha}$$



            Also, remember what $$lim_{xto 0}frac{sin(x)}{x}$$ is equal to?






            share|cite|improve this answer









            $endgroup$



            Hint:



            $$frac{sin(ha)}{h} = acdotfrac{sin(ha)}{ha}$$



            Also, remember what $$lim_{xto 0}frac{sin(x)}{x}$$ is equal to?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 28 at 12:19









            5xum5xum

            91.8k394161




            91.8k394161












            • $begingroup$
              Thanks for show that simple trick!
              $endgroup$
              – VirtualUser
              Jan 28 at 12:45


















            • $begingroup$
              Thanks for show that simple trick!
              $endgroup$
              – VirtualUser
              Jan 28 at 12:45
















            $begingroup$
            Thanks for show that simple trick!
            $endgroup$
            – VirtualUser
            Jan 28 at 12:45




            $begingroup$
            Thanks for show that simple trick!
            $endgroup$
            – VirtualUser
            Jan 28 at 12:45











            0












            $begingroup$

            Notice that $sin(x) = sum_{k = 0}^infty (-1)^k frac{x^{2k + 1}}{(2k + 1)!}$ for all $x in mathbb R$. Thus,
            $$ frac{sin(ah)}{h} = frac 1 h sum_{k = 0}^infty (-1)^k frac{(ah)^{2k + 1}}{(2k + 1)!} = sum_{k = 0}^infty (-1)^k frac{a^{2k + 1} h^{2k}}{(2k + 1)!} longrightarrow a$$
            as $h to 0$ by dominated convergence for example.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Notice that $sin(x) = sum_{k = 0}^infty (-1)^k frac{x^{2k + 1}}{(2k + 1)!}$ for all $x in mathbb R$. Thus,
              $$ frac{sin(ah)}{h} = frac 1 h sum_{k = 0}^infty (-1)^k frac{(ah)^{2k + 1}}{(2k + 1)!} = sum_{k = 0}^infty (-1)^k frac{a^{2k + 1} h^{2k}}{(2k + 1)!} longrightarrow a$$
              as $h to 0$ by dominated convergence for example.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Notice that $sin(x) = sum_{k = 0}^infty (-1)^k frac{x^{2k + 1}}{(2k + 1)!}$ for all $x in mathbb R$. Thus,
                $$ frac{sin(ah)}{h} = frac 1 h sum_{k = 0}^infty (-1)^k frac{(ah)^{2k + 1}}{(2k + 1)!} = sum_{k = 0}^infty (-1)^k frac{a^{2k + 1} h^{2k}}{(2k + 1)!} longrightarrow a$$
                as $h to 0$ by dominated convergence for example.






                share|cite|improve this answer









                $endgroup$



                Notice that $sin(x) = sum_{k = 0}^infty (-1)^k frac{x^{2k + 1}}{(2k + 1)!}$ for all $x in mathbb R$. Thus,
                $$ frac{sin(ah)}{h} = frac 1 h sum_{k = 0}^infty (-1)^k frac{(ah)^{2k + 1}}{(2k + 1)!} = sum_{k = 0}^infty (-1)^k frac{a^{2k + 1} h^{2k}}{(2k + 1)!} longrightarrow a$$
                as $h to 0$ by dominated convergence for example.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 12:32









                YaddleYaddle

                3,136829




                3,136829






























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