let 'f' be a continuous function on 'R' such thatf(1/2n)=(sin(en))e-n^2+2[n]2/n2+[n]+1 where [.] denotes greatest integer function. Then find the value of f(0).

Dear Paidepelly

f(1/2ⁿ)=(sin(eⁿ))e^{-n^2}+2[n]² /(n²+[n]+1)

for f(0) take limit n tends to infinity both side

Lt _n→∞ f(1/2ⁿ)=Lt _n→∞(sin(eⁿ))e^{-n^2}+2[n]² /(n²+[n]+1)

f(0)=Lt _n→∞(sin(eⁿ))e^{-n^2}+2[n]² /(n²+[n]+1)

        =Lt _n→∞(sin(eⁿ))e^{-n^2}+ Lt _n→∞   2[n]² /(n²+[n]+1)

        =0  +Lt _n→∞   2[n]² /(n²+[n]+1)

let n=[n] +f wher f is fractional part of n

so f(0) =Lt _n→∞   2(n-f)² /(n²+n-f+1)

find limit

f(0) =2

Approved