Question 1195866
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                    Step by step.



<pre>
(1)  Write several starting terms of the first series

         cos(x) = 1 - {{{x^2/2}}} + {{{x^4/24}}} - {{{x^6/720}}} .


     Replace here x with {{{x^3}}}, since you want  {{{cos(x^3)}}}.  You will get

         {{{cos(x^3)}}} = 1 - {{{x^6/2}}} + {{{x^12/24}}} - {{{x^18/720}}}.           (1)


     You should keep as many terms of this series to have at least the term with {{{x^12}}}.



(2)  Write several starting terms of the second series

         {{{tan^(-1)(x)}}} = {{{x}}} - {{{x^3/3}}} + {{{x^5/5}}} .


     Replace here x with {{{x^6/2}}}, since you want  {{{tan^(-1)(x^6/2)}}}.  You will get

         {{{tan^(-1)(x^6/2)}}} = {{{x^6/2}}} - {{{x^18/24}}} + {{{x^30/160}}}.          (2)


     You should keep as many terms of this series to have at least the term with {{{x^12}}}.



(3)  Using (1) and (2), form   {{{cos(x^3)}}} + {{{tan^(-1)(x^6/2)}}} - {{{1}}}.   You will get

         {{{cos(x^3)}}} + {{{tan^(-1)(x^6/2)}}} - {{{1}}} =  1 - {{{x^6/2}}} + {{{x^12/24}}} - {{{x^18/720}}} 

                                             +  {{{x^6/2}}} - {{{x^18/24}}} + {{{x^30/160}}} - 1 =

         = some terms will cancel; other will remain; I will keep the remaining terms with {{{x^12}}}.  It gives 

         = {{{x^12/24}}}.   The other terms have x in degrees HIGHER than 12.



(4)  After dividing by {{{x^12}}},  I have {{{1/24}}} 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 <U>ANSWER</U>:  the sough limit at  x --> 0  equals  {{{1/24}}}.
</pre>

Solved.


Is everything clear to you ?