Question

How to solve this?

Answer 1

Answer is (2) i.e. `1/sqrt2`

Explanation:

`lim_(x->3)(sqrt(3x)-3)/((sqrt(2x-4)-sqrt2)`

As when `x->3`, both numerator and denominator approach `0`, we can use L'ospital's rule and hence the limit is equal to

`lim_(x->3)((d(sqrt(3x)-3))/(dx))/((d(sqrt(2x-4)-sqrt2))/(dx))`

= `lim_(x->3)(sqrt3/(2sqrtx))/(2/(2sqrt(2x-4)))`

= `(sqrt3/(2sqrt3))/(2/(2sqrt(2*3-4)))=sqrt3/(2sqrt3)xx(2sqrt(6-4))/2`

= `sqrt2/2`

= `1/sqrt2`

Hence answer is (2).