IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - limits(A-15)

uday_zingtudor

_{[n][infinity]} { 1/2(tan x/2) + 1/2² (tan x/2²) + 1/2³(tan x/2³).......1/2ⁿ(tan x/2ⁿ)}

Adding and subtracting 1/2ⁿcotx/2ⁿ

Now,

We know that cotx-tanx=2cot2x

Here,

- 1/2ⁿcotx/2ⁿ+1/2ⁿtanx/2ⁿ = -1/2^n-1cotx/2^n-1

Doing thus all the way in the problem, we get left with

_{[ n]}

_{[infinity ]} - cotx +1/2ⁿcotx/2ⁿ

_{= - cotx +}_{[ n]}

_{[ infinity]} 1/2ⁿcotx/2ⁿ

^{= - cotx + 1/x { On applying the limits}}

^{= 1/x - cotx}

^{~Cheerio!!!!!}