Finding $lim_{xtopi/2}left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)$ using algebra of limits












4














Let $$
F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $pi/2$.



How can we find the limit using algebra of limits?



The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?



The limit of $F(x)$ is $-frac{3pi^2}{16}$.



How to apply algebra of limits here?






calculus







share|cite|improve this question





























    4














    Let $$
    F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
    $$
    Then find the limit of $F(x)$ as $x$ tends to $pi/2$.



    How can we find the limit using algebra of limits?



    The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?



    The limit of $F(x)$ is $-frac{3pi^2}{16}$.



    How to apply algebra of limits here?






    calculus







    share|cite|improve this question



























      4












      4








      4







      Let $$
      F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
      $$
      Then find the limit of $F(x)$ as $x$ tends to $pi/2$.



      How can we find the limit using algebra of limits?



      The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?



      The limit of $F(x)$ is $-frac{3pi^2}{16}$.



      How to apply algebra of limits here?






      calculus







      share|cite|improve this question















      Let $$
      F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
      $$
      Then find the limit of $F(x)$ as $x$ tends to $pi/2$.



      How can we find the limit using algebra of limits?



      The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?



      The limit of $F(x)$ is $-frac{3pi^2}{16}$.



      How to apply algebra of limits here?






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 '18 at 8:26









      Blue

      47.6k870151




      47.6k870151










      asked Nov 26 '18 at 7:37









      Mathsaddict

      2458




      2458






















          3 Answers
          3






          active

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          3














          The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.



          Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
          $$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
          Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?



          Can you take it from here?






          share|cite|improve this answer































            3














            Make life simpler using $x=y+frac pi 2$ which makes
            $$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
            $$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.






            share|cite|improve this answer





























              1














              Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.



              To solve the limit let $y=pi/2-xto 0$ then



              $$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$



              $$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$






              share|cite|improve this answer























              • I didn't get this indeterminate form. Can you please explain.
                – Mathsaddict
                Nov 26 '18 at 7:47










              • What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                – gimusi
                Nov 26 '18 at 7:54










              • Oh, it is infinity. I made mistake in finding this one. Thanks.
                – Mathsaddict
                Nov 26 '18 at 8:16











              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3














              The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.



              Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
              $$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
              Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?



              Can you take it from here?






              share|cite|improve this answer




























                3














                The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.



                Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
                $$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
                Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?



                Can you take it from here?






                share|cite|improve this answer


























                  3












                  3








                  3






                  The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.



                  Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
                  $$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
                  Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?



                  Can you take it from here?






                  share|cite|improve this answer














                  The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.



                  Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
                  $$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
                  Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?



                  Can you take it from here?







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 26 '18 at 8:19

























                  answered Nov 26 '18 at 8:03









                  Robert Z

                  93.4k1061132




                  93.4k1061132























                      3














                      Make life simpler using $x=y+frac pi 2$ which makes
                      $$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
                      $$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.






                      share|cite|improve this answer


























                        3














                        Make life simpler using $x=y+frac pi 2$ which makes
                        $$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
                        $$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.






                        share|cite|improve this answer
























                          3












                          3








                          3






                          Make life simpler using $x=y+frac pi 2$ which makes
                          $$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
                          $$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.






                          share|cite|improve this answer












                          Make life simpler using $x=y+frac pi 2$ which makes
                          $$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
                          $$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 26 '18 at 8:42









                          Claude Leibovici

                          119k1157132




                          119k1157132























                              1














                              Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.



                              To solve the limit let $y=pi/2-xto 0$ then



                              $$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$



                              $$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$






                              share|cite|improve this answer























                              • I didn't get this indeterminate form. Can you please explain.
                                – Mathsaddict
                                Nov 26 '18 at 7:47










                              • What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                                – gimusi
                                Nov 26 '18 at 7:54










                              • Oh, it is infinity. I made mistake in finding this one. Thanks.
                                – Mathsaddict
                                Nov 26 '18 at 8:16
















                              1














                              Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.



                              To solve the limit let $y=pi/2-xto 0$ then



                              $$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$



                              $$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$






                              share|cite|improve this answer























                              • I didn't get this indeterminate form. Can you please explain.
                                – Mathsaddict
                                Nov 26 '18 at 7:47










                              • What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                                – gimusi
                                Nov 26 '18 at 7:54










                              • Oh, it is infinity. I made mistake in finding this one. Thanks.
                                – Mathsaddict
                                Nov 26 '18 at 8:16














                              1












                              1








                              1






                              Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.



                              To solve the limit let $y=pi/2-xto 0$ then



                              $$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$



                              $$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$






                              share|cite|improve this answer














                              Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.



                              To solve the limit let $y=pi/2-xto 0$ then



                              $$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$



                              $$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 26 '18 at 7:59

























                              answered Nov 26 '18 at 7:41









                              gimusi

                              1




                              1












                              • I didn't get this indeterminate form. Can you please explain.
                                – Mathsaddict
                                Nov 26 '18 at 7:47










                              • What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                                – gimusi
                                Nov 26 '18 at 7:54










                              • Oh, it is infinity. I made mistake in finding this one. Thanks.
                                – Mathsaddict
                                Nov 26 '18 at 8:16


















                              • I didn't get this indeterminate form. Can you please explain.
                                – Mathsaddict
                                Nov 26 '18 at 7:47










                              • What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                                – gimusi
                                Nov 26 '18 at 7:54










                              • Oh, it is infinity. I made mistake in finding this one. Thanks.
                                – Mathsaddict
                                Nov 26 '18 at 8:16
















                              I didn't get this indeterminate form. Can you please explain.
                              – Mathsaddict
                              Nov 26 '18 at 7:47




                              I didn't get this indeterminate form. Can you please explain.
                              – Mathsaddict
                              Nov 26 '18 at 7:47












                              What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                              – gimusi
                              Nov 26 '18 at 7:54




                              What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
                              – gimusi
                              Nov 26 '18 at 7:54












                              Oh, it is infinity. I made mistake in finding this one. Thanks.
                              – Mathsaddict
                              Nov 26 '18 at 8:16




                              Oh, it is infinity. I made mistake in finding this one. Thanks.
                              – Mathsaddict
                              Nov 26 '18 at 8:16


















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