Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=4-1/(x+5) +1/x`?

Answer 1

No removanble discontinues, vertical asymptotes at `x=0` and `x=-5` and horizontal asymptotes at `y=4`

Explanation:

As `f(x)=4-1/(x+5)+1/x=(4x(x+5)-x+x+5)/(x(x+5))`

= `(4x^2+20x+5)/(x(x+5)`

As `x` or `x+5` is not a factor of `4x^2+20x+5`, there are no removanble discontinues.

Vertical asymptotes are at `x=0` and `x+5=0` i.e. `x=-5`, because as `x->0` or `x->-5`, `f(x)->+-oo`, depending on whether we approach from left or right.

Now we can write `f(x)=(4x^2+20x+5)/(x(x+5)`

= `(4x^2+20x+5)/(x^2+5x)`

= `(4+20/x+5/x^2)/(1+5/x)`

Hence as `x->oo`, `f(x)->4`

and we have horizontal asymptote `y=4`

graph{4-1/(x+5)+1/x [-21.92, 18.08, -5.08, 14.92]}