Question

How do you find `lim (5x+6)/(x^2-4)` as `x->oo`?

Answer 1

The answer is `=0^-` and `0^+`

Explanation:

`lim_(x->-oo)(5x+6)/(x^2-4)=lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))`

`lim_(x->-oo)6/x=0`

`lim_(x->-oo)4/x^2=0`

`lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->-oo)5/x=0^-`

`lim_(x->+oo)(5x+6)/(x^2-4)=lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))`

`lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->+oo)5/x=0^+`