Compute the limit:












0














$$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
I tried:
$$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
$$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
$$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
Somewhere I over-complicated and I don't know how to continue or the other route that I should take...






limits trigonometry







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    0














    $$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
    I tried:
    $$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
    $$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
    $$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
    $$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
    Somewhere I over-complicated and I don't know how to continue or the other route that I should take...






    limits trigonometry







    share|cite|improve this question



























      0












      0








      0







      $$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
      I tried:
      $$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
      $$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
      $$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
      $$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
      Somewhere I over-complicated and I don't know how to continue or the other route that I should take...






      limits trigonometry







      share|cite|improve this question















      $$ lim_{x geq 0}{frac{1-cos^3(x)}{xsin(2x)}}$$
      I tried:
      $$lim_{x geq 0}{{frac{1-cos^2(x)*cos(x)}{2*x*sin^2(x)cos^2(x)}}}$$
      $$lim_{x geq 0}{frac{cos^2x+sin^2x-cos^2(x)cos(x)}{2*x*sin^2x*cos^2(x)}}$$
      $$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-(1-sin^2(x))cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
      $$lim_{x geq 0}{frac{cos^2(x)+sin^2(x)-cos(x)+sin^2(x)cos(x)}{2*x*sin^2(x)*cos^2(x)}}$$
      Somewhere I over-complicated and I don't know how to continue or the other route that I should take...






      limits trigonometry




      limits trigonometry






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      edited Dec 9 at 0:48









      Key Flex

      7,46841232




      7,46841232










      asked Dec 8 at 16:27









      Bili Debili

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          5 Answers
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          2














          $$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
          Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
          Again apply L'Hopital's Rule
          $$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$






          share|cite|improve this answer





























            1














            Hint: Write
            $$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$






            share|cite|improve this answer





























              1














              begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
              &= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
              &= lim_{x to 0} frac{frac32x^2}{2x^2}\
              &= frac34end{align}






              share|cite|improve this answer





















              • What is the step in between the first computation, where you get rid of trigonometric identities
                – Bili Debili
                Dec 8 at 16:44












              • taylor series, just drop the higher order terms.
                – Siong Thye Goh
                Dec 8 at 16:47






              • 1




                I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                – gimusi
                Dec 8 at 17:00



















              1














              We have that



              $${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$



              then refer to standard limits




              • $frac{1-cos(x)}{x^2}to frac12$

              • $frac{2x}{sin(2x)}to 1$


              and simply




              • $1+cos x+cos^2(x)to 3$






              share|cite|improve this answer































                0














                Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.






                share|cite|improve this answer





















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                  5 Answers
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                  5 Answers
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                  2














                  $$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
                  Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
                  Again apply L'Hopital's Rule
                  $$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$






                  share|cite|improve this answer


























                    2














                    $$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
                    Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
                    Again apply L'Hopital's Rule
                    $$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$






                    share|cite|improve this answer
























                      2












                      2








                      2






                      $$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
                      Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
                      Again apply L'Hopital's Rule
                      $$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$






                      share|cite|improve this answer












                      $$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}$$
                      Apply L'Hopital's Rule$$lim_{xrightarrow0}dfrac{1-cos^3x}{xsin(2x)}=lim_{xrightarrow0}dfrac{3cos^2xsin x}{sin(2x)+2xcos(2x)}$$
                      Again apply L'Hopital's Rule
                      $$lim_{xrightarrow0}dfrac{3(-sin(2x)sin x+cos^3x)}{4cos(2x)-4xsin(2x)}=dfrac34$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 8 at 16:41









                      Key Flex

                      7,46841232




                      7,46841232























                          1














                          Hint: Write
                          $$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$






                          share|cite|improve this answer


























                            1














                            Hint: Write
                            $$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$






                            share|cite|improve this answer
























                              1












                              1








                              1






                              Hint: Write
                              $$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$






                              share|cite|improve this answer












                              Hint: Write
                              $$frac{(1-cos(x))(1+cos(x))(cos^2(x)+cos(x)+1)}{2xsin(x)cos(x)(1+cos(x))}$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 8 at 16:33









                              Dr. Sonnhard Graubner

                              72.8k32865




                              72.8k32865























                                  1














                                  begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
                                  &= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
                                  &= lim_{x to 0} frac{frac32x^2}{2x^2}\
                                  &= frac34end{align}






                                  share|cite|improve this answer





















                                  • What is the step in between the first computation, where you get rid of trigonometric identities
                                    – Bili Debili
                                    Dec 8 at 16:44












                                  • taylor series, just drop the higher order terms.
                                    – Siong Thye Goh
                                    Dec 8 at 16:47






                                  • 1




                                    I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                                    – gimusi
                                    Dec 8 at 17:00
















                                  1














                                  begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
                                  &= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
                                  &= lim_{x to 0} frac{frac32x^2}{2x^2}\
                                  &= frac34end{align}






                                  share|cite|improve this answer





















                                  • What is the step in between the first computation, where you get rid of trigonometric identities
                                    – Bili Debili
                                    Dec 8 at 16:44












                                  • taylor series, just drop the higher order terms.
                                    – Siong Thye Goh
                                    Dec 8 at 16:47






                                  • 1




                                    I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                                    – gimusi
                                    Dec 8 at 17:00














                                  1












                                  1








                                  1






                                  begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
                                  &= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
                                  &= lim_{x to 0} frac{frac32x^2}{2x^2}\
                                  &= frac34end{align}






                                  share|cite|improve this answer












                                  begin{align}lim_{x to 0}frac{1-cos^3x}{x sin (2x)}&= lim_{x to 0}frac{1-left( 1-frac{x^2}2right)^3}{2x^2}\
                                  &= lim_{x to 0} frac{1-(1-frac{3x^2}2)}{2x^2} \
                                  &= lim_{x to 0} frac{frac32x^2}{2x^2}\
                                  &= frac34end{align}







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Dec 8 at 16:37









                                  Siong Thye Goh

                                  98.9k1464116




                                  98.9k1464116












                                  • What is the step in between the first computation, where you get rid of trigonometric identities
                                    – Bili Debili
                                    Dec 8 at 16:44












                                  • taylor series, just drop the higher order terms.
                                    – Siong Thye Goh
                                    Dec 8 at 16:47






                                  • 1




                                    I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                                    – gimusi
                                    Dec 8 at 17:00


















                                  • What is the step in between the first computation, where you get rid of trigonometric identities
                                    – Bili Debili
                                    Dec 8 at 16:44












                                  • taylor series, just drop the higher order terms.
                                    – Siong Thye Goh
                                    Dec 8 at 16:47






                                  • 1




                                    I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                                    – gimusi
                                    Dec 8 at 17:00
















                                  What is the step in between the first computation, where you get rid of trigonometric identities
                                  – Bili Debili
                                  Dec 8 at 16:44






                                  What is the step in between the first computation, where you get rid of trigonometric identities
                                  – Bili Debili
                                  Dec 8 at 16:44














                                  taylor series, just drop the higher order terms.
                                  – Siong Thye Goh
                                  Dec 8 at 16:47




                                  taylor series, just drop the higher order terms.
                                  – Siong Thye Goh
                                  Dec 8 at 16:47




                                  1




                                  1




                                  I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                                  – gimusi
                                  Dec 8 at 17:00




                                  I always suggest to include remainder terms (in little-o or big-O form) when using Taylor's expansion, especialy when we introduce to the method not expert student.
                                  – gimusi
                                  Dec 8 at 17:00











                                  1














                                  We have that



                                  $${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$



                                  then refer to standard limits




                                  • $frac{1-cos(x)}{x^2}to frac12$

                                  • $frac{2x}{sin(2x)}to 1$


                                  and simply




                                  • $1+cos x+cos^2(x)to 3$






                                  share|cite|improve this answer




























                                    1














                                    We have that



                                    $${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$



                                    then refer to standard limits




                                    • $frac{1-cos(x)}{x^2}to frac12$

                                    • $frac{2x}{sin(2x)}to 1$


                                    and simply




                                    • $1+cos x+cos^2(x)to 3$






                                    share|cite|improve this answer


























                                      1












                                      1








                                      1






                                      We have that



                                      $${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$



                                      then refer to standard limits




                                      • $frac{1-cos(x)}{x^2}to frac12$

                                      • $frac{2x}{sin(2x)}to 1$


                                      and simply




                                      • $1+cos x+cos^2(x)to 3$






                                      share|cite|improve this answer














                                      We have that



                                      $${frac{1-cos^3(x)}{xsin(2x)}}={frac{1-cos(x)}{x^2}}cdot(1+cos x+cos^2(x))cdotfrac12{frac{2x}{sin(2x)}}$$



                                      then refer to standard limits




                                      • $frac{1-cos(x)}{x^2}to frac12$

                                      • $frac{2x}{sin(2x)}to 1$


                                      and simply




                                      • $1+cos x+cos^2(x)to 3$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 8 at 16:59

























                                      answered Dec 8 at 16:30









                                      gimusi

                                      1




                                      1























                                          0














                                          Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.






                                          share|cite|improve this answer


























                                            0














                                            Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.






                                            share|cite|improve this answer
























                                              0












                                              0








                                              0






                                              Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.






                                              share|cite|improve this answer












                                              Let's write the expression under limit as $$frac{1-cos ^3x}{1-cos x} cdotfrac{1-cos x} {x^2}cdotfrac{2x}{sin 2x}cdotfrac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 9 at 5:06









                                              Paramanand Singh

                                              48.8k555158




                                              48.8k555158






























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                                                  VLT Carioca

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                                                  VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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