Question

How do you find the limit of `sqrt(x^2-9)/(2x-6)` as `x->oo`?

Answer 1

`limx->oo[(sqrt(x^2-9))/(2x-6)]`
`= 1/2`

Explanation:

`limx->oo[(sqrt(x^2-9))/(2x-6)]=oo/oo` using direct substitution
So, we can use L'Hospital's rule:
`limx->oo[((2x)/(2sqrt(x^2-9)))/(2)]`
This simplifies to:
`limx->oo[x/(2sqrt(x^2-9))]`
Use L'Hospital's rule again:
`limx->oo[1/[((2x)/sqrt(x^2-9))]]`
`= limx->oo[sqrt(x^2-9)/(2x)]`
`=1/2`