Question

How do I determine the end behavior of the graph, f(x)=(3x-3)/(4x+5), in limit notation?

Answer 1

See below.

Explanation:

The end behaviour is what is happening as `x->+-oo`

`lim_(x->oo)(3x-3)/(4x+5)`

`(3x-3)/(4x+5)`

Divide by `x`:

`((3x)/x-3/x)/((4x)/x+5/x)=(3-3/x)/(4+5/x)`

`lim_(x->oo)(3x-3)/(4x+5)=>(3-3/x)/(4+5/x)->(3-0)/(4+0)=3/4`

As before:

`lim_(x->-oo)(3x-3)/(4x+5)=>(3-3/x)/(4+5/x)->(3-0)/(4+0)=3/4`

This shows that the line `y=3/4` is a horizontal asymptote.

So as `x->+-oo` the function tends to `3/4`.

The graph of `f(x)` confirms this: