Friday, 20 January 2017

Limit involving arctan without L'Hopital rule and series expansions



Find limx0+parctan(x/p)qarctan(x/q)xx

without L'Hopital rule and series expansions.



Could someone help me , I did not understand how to find it, thanks.


Answer




Recall arctany=y01/(1+t2)dt. Verify that



1t211+t21t2+t4



for all t. Thus for y>0,



yy3/3arctanyyy3/3+y5/5.



It follow that our expression is bounded below by




p(x/px3/2/(3p3))q(x/qx3/2/(3q3)+x5/2/(5q5))x3/2.



Simplify to see this 1/(3q2)1/(3p2). There is a similar estimate from above, giving the same limit. By the squeeze theorem, the limit is 1/(3q2)1/(3p2).


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