Question

Let `p=(lim)_(xvec0+)(1+tan^2sqrt(x))^(1//2x)` then `logp` is equal to: (1) 2 (2) 1 (3) `1/2` (4) `1/4`

Answer 1

`p = lim_(x->0^+) ( 1+ tan^2 sqrtx)^(1/(2x))`
`log p = lim_(x->0^+) log(1+ tan^2 sqrtx)^(1/(2x))`
`= lim_(x->0^+) 1/(2x) log(1+ tan^2 sqrtx)`
`= lim_(x->0^+) (Log(1+tan^2 sqrt x))/(2x)`
using L hospital rule
`lim_(x->0) (ax^2)/(bx^3) = (2ax)/(3bx^2)`
`= lim_(x->0^+) (1/(1+ tan^2 sqrtx) xx 2 tan sqrt xx sec^2 sqrt x xx 1/(2 sqrt x))/2`
`log p = lim_(x->0+) ( tan sqrt x xx sec^2 sqrt x)/ ( 2 sqrt x xx ( 1 + tan^2 sqrtx))`
`= (1 xx 1)/(2 xx (1+0))`
`log p = 1/2 `
option 2 is correct Answer