`("lim")_(xvecoo)"{"x+5")"tan^(-1)(x+5)-(x+1)tan^(-1)(x+1)}` is equal to (a) `pi` (b) `2pi` (c) `pi/2` (d) none of these

Options

A. `(1)/(m2^(m))-(1)/(n2^(n))`
B. `(1)/(m2^(m))+(1)/(n2^(n))`
C. `(1)/(m2^(m-1))-(1)/(n2^(n-1))`
D. `(1)/(m2^(m-1))+(1)/(n2^(n-1))`

Answer

B

Explanation

Given limit is
`underset(xtooo)lim{(x+1)[tan^(-1)(x+5)-(x+1)]+4tan^(-1)(x+5)`
`=underset(xtooo)lim[(x+1)"tan"^(-1)(4)/(1+(x+1)(x+5))+4tan^(-1)(x+5)]`
`=underset(xtooo)lim[(x+1)("tan"^(-1)(4)/(x^(2)+6x+6))/(((4)/(x^(2)+6x+6)))xx(4)/(x^(2)+6x+6)+4tan^(-1)(x+5)]`
`=0+4xx(pi)/(2)=2pi`