Question

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

Answer 1

`lim_(x->oo)sin((x-1)/(2+x^2))=0`

Explanation:

Because `f(x)=sin(x)` is continuous, we have

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

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

`=sin((0-0)/(0+1))`

`=sin(0)`

`=0`