Question

How do you find the limit of `sin(x-1)/(x^2+x-2)` as `x->1`?

Answer 1

`1/3`

Explanation:

`sin(x-1)/(x^2+x-2)= sin(x-1)/((x+2)(x-1)) = (sin(x-1)/(x-1)) 1/(x+2)`

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