If `f : [-1, 1] to R and f'(0) = lim_( n to infty) nf(1/n) and f(0) = 0`. Find the value of `lim_( n to infty) 2/pi (n+1) cos^(-1) (1/n) -n`, given that `0 lt| lim_( n to infty) cos^(-1)(1/n)| lt pi/2.`

Answer

`(1-2/pi)`

Explanation

Here, `underset( n to infty) lim 2/pi (n+1) cos^(-1) (1/n) -n `
` underset( n to infty) n{2/pi (1+1/n) cos^(-1) (1/n) -1} = underset( n to infty) n f(1/n)`
where, ` f(1/n) = 2/pi (1+1/n) cos^(-1) (1/n) -1 = f'(0)`
`["given, " f'(0) = underset( n to infty) nf(1/n)]`
`:." "underset( n to infty) lim 2/pi (n+1) cos^(-1) 1/n -n = f'(0) `...(i)
where, ` f(x) = 2/pi (1+x) cos^(-1) x -1, f(0) = 0`
`rArr" " f'(x) = 2/pi {(1+x)(-1)/sqrt(1-x^(2)) + cos^(-1) x}`
` rArr" " f'(0) = 2/pi {-1+pi/2} = 1-2/pi` ...(ii)
`:.` From Eqs. (i) and (ii), we get
` underset( n to infty) lim 2/pi (n+1) cos^(-1) (1/n) -n = 1 - 2/pi`