Evaluate: `underset(nrarrinfty)lim[(1/n^2sec^2(1/(n^2))+2/n^2sec^2((4)/(n^2))....+1/nsec^2(1)]`

Answer

`1/2 tan1`

Explanation

`lim_(n to oo) [1/(n^(2)) "sec"^(2)1/(n^(2))+2/(n^(2))"sec"^(2) 4/(n^(2))+……..+1/nsec^(2)1]`
`=lim_(nto oo) 1/n sum_(r=1)^(n)r/n"sec"^(2)(r/n)^(2)`
`=int_(0)^(1)x sec^(2) x^(2) dx`
Put `x^(2)=t` so that `2x dx=dt`
When `x=0, t=0`, When `x=1, t=1`
`:. `Required limit `=1/2 int_(0)^(1) sec^(2) t dt `
`=1/2 [tan t]_(0)^(1)=1/2[tan 1-0]`
`=1/2 tan 1`