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limx->0(1/x√x) (a arc tan√x/a -b arc tan √x/b) has the value equal to

Tanya , 6 Years ago
Grade 11
anser 1 Answers
Aditya Gupta

Last Activity: 6 Years ago

let rootx=y
then we have
limx->0(1/x√x) (a arc tan√x/a -b arc tan √x/b) = limy->0(1/y^3) (a arc tany/a -b arc tany/b) 
now use series expansion of arctanz= z – z^3/3+z^5/5 – …...
so limit is
limy->0(1/y^3)[a(y/a – y^3/3a^3+...) – b(y/b – y^3/3b^3+.....)]
= (a^2 – b^2)/3a^2b^2

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