Evaluating $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{sin3theta } } $












1














$$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{sin3theta } } $$



My impression:



I don't see how I can change the variables in any way to get it in the form where I can simplify it to 1. I also looked up similar problems and they mention L'Hopital's rule to solve this, which I haven't learned yet. Do not give a straight answer. I already have that available to me.






calculus limits







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  • Hint: Break it up into the product of two limits
    – ASKASK
    Jan 20 '15 at 4:07
















1














$$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{sin3theta } } $$



My impression:



I don't see how I can change the variables in any way to get it in the form where I can simplify it to 1. I also looked up similar problems and they mention L'Hopital's rule to solve this, which I haven't learned yet. Do not give a straight answer. I already have that available to me.






calculus limits







share|cite|improve this question
























  • Hint: Break it up into the product of two limits
    – ASKASK
    Jan 20 '15 at 4:07














1












1








1







$$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{sin3theta } } $$



My impression:



I don't see how I can change the variables in any way to get it in the form where I can simplify it to 1. I also looked up similar problems and they mention L'Hopital's rule to solve this, which I haven't learned yet. Do not give a straight answer. I already have that available to me.






calculus limits







share|cite|improve this question















$$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{sin3theta } } $$



My impression:



I don't see how I can change the variables in any way to get it in the form where I can simplify it to 1. I also looked up similar problems and they mention L'Hopital's rule to solve this, which I haven't learned yet. Do not give a straight answer. I already have that available to me.






calculus limits




calculus limits






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share|cite|improve this question













share|cite|improve this question




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edited Nov 23 at 17:32









Martin Sleziak

44.6k7115270




44.6k7115270










asked Jan 20 '15 at 4:06









Cherry_Developer

1,93821431




1,93821431












  • Hint: Break it up into the product of two limits
    – ASKASK
    Jan 20 '15 at 4:07


















  • Hint: Break it up into the product of two limits
    – ASKASK
    Jan 20 '15 at 4:07
















Hint: Break it up into the product of two limits
– ASKASK
Jan 20 '15 at 4:07




Hint: Break it up into the product of two limits
– ASKASK
Jan 20 '15 at 4:07










2 Answers
2






active

oldest

votes


















4














Hint:



$$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } } = lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } cdot frac{theta}{theta}} = lim_{theta rightarrow 0}{frac{sin(7theta)}{theta} } cdot {frac{theta}{sin(3theta)} }$$



And then do each limit independently.






share|cite|improve this answer























  • I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
    – Cherry_Developer
    Jan 20 '15 at 4:12










  • Sorry, I'll make an edit
    – ASKASK
    Jan 20 '15 at 4:13










  • Basically all I did was multiply the top and bottom by $theta$
    – ASKASK
    Jan 20 '15 at 4:15










  • Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
    – Cherry_Developer
    Jan 20 '15 at 4:17










  • It went to the otehr limit
    – ASKASK
    Jan 20 '15 at 4:18



















4














If you're allowed to use Taylor series, note that $sin(ktheta) approx ktheta$ when $theta$ is near zero:



Solution using this technique, now that I see you probably have solved this:




$$lim_{theta to 0} frac{sin(7theta)}{sin(3theta)} = lim_{thetato0}frac{7theta}{3theta} = frac{7}{3}$$







share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    Hint:



    $$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } } = lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } cdot frac{theta}{theta}} = lim_{theta rightarrow 0}{frac{sin(7theta)}{theta} } cdot {frac{theta}{sin(3theta)} }$$



    And then do each limit independently.






    share|cite|improve this answer























    • I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
      – Cherry_Developer
      Jan 20 '15 at 4:12










    • Sorry, I'll make an edit
      – ASKASK
      Jan 20 '15 at 4:13










    • Basically all I did was multiply the top and bottom by $theta$
      – ASKASK
      Jan 20 '15 at 4:15










    • Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
      – Cherry_Developer
      Jan 20 '15 at 4:17










    • It went to the otehr limit
      – ASKASK
      Jan 20 '15 at 4:18
















    4














    Hint:



    $$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } } = lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } cdot frac{theta}{theta}} = lim_{theta rightarrow 0}{frac{sin(7theta)}{theta} } cdot {frac{theta}{sin(3theta)} }$$



    And then do each limit independently.






    share|cite|improve this answer























    • I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
      – Cherry_Developer
      Jan 20 '15 at 4:12










    • Sorry, I'll make an edit
      – ASKASK
      Jan 20 '15 at 4:13










    • Basically all I did was multiply the top and bottom by $theta$
      – ASKASK
      Jan 20 '15 at 4:15










    • Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
      – Cherry_Developer
      Jan 20 '15 at 4:17










    • It went to the otehr limit
      – ASKASK
      Jan 20 '15 at 4:18














    4












    4








    4






    Hint:



    $$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } } = lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } cdot frac{theta}{theta}} = lim_{theta rightarrow 0}{frac{sin(7theta)}{theta} } cdot {frac{theta}{sin(3theta)} }$$



    And then do each limit independently.






    share|cite|improve this answer














    Hint:



    $$lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } } = lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ sin3theta } cdot frac{theta}{theta}} = lim_{theta rightarrow 0}{frac{sin(7theta)}{theta} } cdot {frac{theta}{sin(3theta)} }$$



    And then do each limit independently.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 20 '15 at 4:15

























    answered Jan 20 '15 at 4:10









    ASKASK

    5,87431836




    5,87431836












    • I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
      – Cherry_Developer
      Jan 20 '15 at 4:12










    • Sorry, I'll make an edit
      – ASKASK
      Jan 20 '15 at 4:13










    • Basically all I did was multiply the top and bottom by $theta$
      – ASKASK
      Jan 20 '15 at 4:15










    • Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
      – Cherry_Developer
      Jan 20 '15 at 4:17










    • It went to the otehr limit
      – ASKASK
      Jan 20 '15 at 4:18


















    • I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
      – Cherry_Developer
      Jan 20 '15 at 4:12










    • Sorry, I'll make an edit
      – ASKASK
      Jan 20 '15 at 4:13










    • Basically all I did was multiply the top and bottom by $theta$
      – ASKASK
      Jan 20 '15 at 4:15










    • Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
      – Cherry_Developer
      Jan 20 '15 at 4:17










    • It went to the otehr limit
      – ASKASK
      Jan 20 '15 at 4:18
















    I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
    – Cherry_Developer
    Jan 20 '15 at 4:12




    I see how to go from here... but wouldn't it become $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot lim _{ theta rightarrow 0 }{ frac { 1 }{ sin3theta } } $ I don't get where you got the $theta$
    – Cherry_Developer
    Jan 20 '15 at 4:12












    Sorry, I'll make an edit
    – ASKASK
    Jan 20 '15 at 4:13




    Sorry, I'll make an edit
    – ASKASK
    Jan 20 '15 at 4:13












    Basically all I did was multiply the top and bottom by $theta$
    – ASKASK
    Jan 20 '15 at 4:15




    Basically all I did was multiply the top and bottom by $theta$
    – ASKASK
    Jan 20 '15 at 4:15












    Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
    – Cherry_Developer
    Jan 20 '15 at 4:17




    Sorry, still confused... $lim _{ theta rightarrow 0 }{ frac { sin { 7theta } }{ 1 } } cdot frac { theta }{ theta } =lim _{ theta rightarrow 0 }{ frac { theta sin { 7theta } }{ theta } } $ no? Where did the $theta$ in the numerator disappear to?
    – Cherry_Developer
    Jan 20 '15 at 4:17












    It went to the otehr limit
    – ASKASK
    Jan 20 '15 at 4:18




    It went to the otehr limit
    – ASKASK
    Jan 20 '15 at 4:18











    4














    If you're allowed to use Taylor series, note that $sin(ktheta) approx ktheta$ when $theta$ is near zero:



    Solution using this technique, now that I see you probably have solved this:




    $$lim_{theta to 0} frac{sin(7theta)}{sin(3theta)} = lim_{thetato0}frac{7theta}{3theta} = frac{7}{3}$$







    share|cite|improve this answer




























      4














      If you're allowed to use Taylor series, note that $sin(ktheta) approx ktheta$ when $theta$ is near zero:



      Solution using this technique, now that I see you probably have solved this:




      $$lim_{theta to 0} frac{sin(7theta)}{sin(3theta)} = lim_{thetato0}frac{7theta}{3theta} = frac{7}{3}$$







      share|cite|improve this answer


























        4












        4








        4






        If you're allowed to use Taylor series, note that $sin(ktheta) approx ktheta$ when $theta$ is near zero:



        Solution using this technique, now that I see you probably have solved this:




        $$lim_{theta to 0} frac{sin(7theta)}{sin(3theta)} = lim_{thetato0}frac{7theta}{3theta} = frac{7}{3}$$







        share|cite|improve this answer














        If you're allowed to use Taylor series, note that $sin(ktheta) approx ktheta$ when $theta$ is near zero:



        Solution using this technique, now that I see you probably have solved this:




        $$lim_{theta to 0} frac{sin(7theta)}{sin(3theta)} = lim_{thetato0}frac{7theta}{3theta} = frac{7}{3}$$








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 20 '15 at 6:12

























        answered Jan 20 '15 at 4:07









        apnorton

        15.1k33696




        15.1k33696






























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