`lim_(x->sqrt(3))(x^2-3)/(x^2+3sqrt(3)x-12)`

Explanation

Given,

`lim_(x->sqrt3)(x^2−3)/(x^2+3sqrt3x−12)`

`=lim_(x->sqrt3)((sqrt3)^2−3)/((sqrt3)^2+3sqrt3(sqrt3)−12)`

`=0/0`

which is `0/0` form

So,applying L-Hospital rule,

`lim_(x->sqrt3)(x^2−3)/(x^2+3sqrt3x−12)`

`=lim_(x->sqrt3)(2x)/(2x+3sqrt3)`

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