`lim_(x->oo)cot^(-1)(x^(-a)log_a x)/(sec^(-1)(a^xlog_x a)),(a >1)`is equal to
(a)`2`
(b)` 1`
(c) `(log)_a2`
(d) 0

Answer

B

Explanation

`underset(xtooo)lim(cot^(-1)(x^(-a)log_(a)x))/(sec^(-1)(a^(x)log_(x)a))" "(agt1)`
`=underset(xtooo)lim(cot^(-1)((log_(a)x)/(x^(a))))/(sec^(-1)(a^(x)/(log_(a)x)))`
as` xtooo,((log_(a)x)/(x^(a)))to0" and "(a^(x)/(log_(a)x))tooo`
(using L'Hospital rule)
`:.I=(pi//2)/(pi//2)=1`