SOLUTION: This is a calculus question. Could you explain #16 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200FE-F2021.pdf which is about finding the limi

Algebra ->  Equations -> SOLUTION: This is a calculus question. Could you explain #16 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200FE-F2021.pdf which is about finding the limi      Log On


   

Question 1199202: This is a calculus question. Could you explain #16 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200FE-F2021.pdf which is about finding the limit using series?
Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
.

                    Step by step. 


(1)  Write several starting terms of the first series

         cos(x) = 1 - x%5E2%2F2 + x%5E4%2F24 - x%5E6%2F720 .


     Replace here x with x%5E3, since you want  cos%28x%5E3%29.  You will get

         cos%28x%5E3%29 = 1 - x%5E6%2F2 + x%5E12%2F24 - x%5E18%2F720.           (1)


     You should keep as many terms of this series to have at least the term with x%5E12.



(2)  Write several starting terms of the second series

         tan%5E%28-1%29%28x%29 = x - x%5E3%2F3 + x%5E5%2F5 .


     Replace here x with x%5E6%2F2, since you want  tan%5E%28-1%29%28x%5E6%2F2%29.  You will get

         tan%5E%28-1%29%28x%5E6%2F2%29 = x%5E6%2F2 - x%5E18%2F24 + x%5E30%2F160.          (2)


     You should keep as many terms of this series to have at least the term with x%5E12.



(3)  Using (1) and (2), form   cos%28x%5E3%29 + tan%5E%28-1%29%28x%5E6%2F2%29 - 1.   You will get

         cos%28x%5E3%29 + tan%5E%28-1%29%28x%5E6%2F2%29 - 1 =  1 - x%5E6%2F2 + x%5E12%2F24 - x%5E18%2F720 

                                             +  x%5E6%2F2 - x%5E18%2F24 + x%5E30%2F160 - 1 =

         = some terms will cancel; other will remain; I will keep the remaining terms with x%5E12.  It gives 

         = x%5E12%2F24.   The other terms have x in degrees HIGHER than 12.



(4)  After dividing by x%5E12,  I have 1%2F24 plus other terms with x of degree higher than 1.

     When calculating the limit at x--> 0, these terms produce 0 (zero), so they are not interesting to me.



(5)  Thus I get the ANSWER:  the sough limit at  x --> 0  equals  1%2F24.

Solved.

Is everything clear to you ?