If `L= lim_(xto2) (root(3)(60+x^(2))-4)/(sin(x-2))`, then the value of `1//L` is________.

Answer

`(12)`

Explanation

We have, `underset(xto2)lim(root(3)(60+x^(2))-root(3)(64))/(sin(x-2))`
`=underset(xto2)lim([(60+x^(2))^((1)/(3))-64^((1)/(3))][(60+x^(2))^((2)/(3))+(60+x^(2))^((1)/(3))64^((1)/(3))64^((2)/(3))])/((x-2)(sin(x-2))/((x-2))[(60+x^(2))^((2)/(3))+(60+x^(2))^((1)/(3))64^((1)/(3))+64^((2)/(3))])`
`=underset(xto2)lim(60+x^(2)-64)/((x-2)[16+4xx4+16])`
`=underset(xto2)lim((x-2)(x+2))/(48(x-2))`
`=underset(xto2)lim(x+2)/(48)=(4)/(48)=(1)/(12)`