Question

Question #3721e

Answer 1

`4/3`

Explanation:

Making `y = (x-1)` we have

`lim_(x->1)=(sin^2(x^2-1)tan^2(x-1))/((x^3-1)sin((x-1)^3)) =`

`= lim_(y->0)=(sin^2(y(x+1))tan^2y)/(y(x^2+x+1)sin(y^3)) =`

`= lim_(y->0)(sin^2(y(x+1))sin^2y)/(y^4(x^2+x+1) cos^2 y(sin(y^3)/y^3))=`

now considering that `lim_(y->0)cos^2y = 1`

`= lim_(y->0)(x+1)^2/(x^2+x+1)(sin^2(y(x+1))/(y(x+1))^2sin^2/y^2)/(sin(y^3)/y^3)=4/3`