Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)= (x^2+1)/(x+1)`?

Answer 1

A vertical asymptote is `x=-1`
And the oblique asymptote is `y=x-1`

Explanation:

As `f(x)=(x^2+1)/(x+1)`
Since we cannot divide by `0`
The degree of the binomial of the numerator `>`the degree of the binomial of the denominator, so we would expect an oblique asymptote
Let 's do a long division

`x^2` `color(white)(aaaaaaaa)` `+1` `color(white)(aaaaaa)` `∣` `x+1`
`x^2+x` `color(white)(aaaaaaaaaaaaa)` `∣` `x-1`
`color(white)(aaaa)` `0-x+1`
`color(white)(aaaaaa)` `-x-1`
`color(white)(aaaaaaaa)` `0+2`

And finally we get
`f(x)=x-1+2/(x+1)`
So the oblique asymptote is `y=x-1`

`lim f(x)=-oo`
`x->-oo`

`lim f(x)=+oo`
`x->+oo`