Question

What are the asymptote(s) and hole(s), if any, of `f(x) =(1+1/x)/x`?

Answer 1

The first step is always to simplify the complex fraction to see if we may obtain the function in the form `f(x) = (g(x))/(h(x))`.

`f(x) = ((x + 1)/x)/x`

`f(x) = (x + 1)/x^2`

There will be vertical asymptotes whenever the denominator equals `0`.

`x^2 = 0 -> x = 0`

There is therefore only one vertical asymptote at `x= 0`.

Now we calculate the horizontal asymptotes. I will write an answer that doesn't involve calculus and then an answer that does involve calculus.

No Calculus

By the rules of asymptotes, there will be a horizontal asymptote at `y = 0` (because the highest degree of the denominator is higher than the numerator).

Calculus

We divide each term by the highest degree of the entire function, which will be `x^2` and take the limit as `x` approaches positive infinity.

`y = lim_(x-> oo) (x/x^2 + 1/x^2)/(x^2/x^2)`

`y = lim_(x-> oo) (1/x + 1/x^2)/1`

By the identity `lim_(x-> oo) 1/x = 0`, we have

`y = (0 + 0)/1`

`y = 0`

This confirms what we found above without any calculus.

Practice exercises

`1`. Describe asymptotes, both horizontal and vertical, in the following function.

`g(x) = (x + 3 + 2/(x - 4))/(x - 1)`

`2`. Solutions:

`1`. V.A: `x = 4 and x = 1`. H.A: `y = 1`

Hopefully this helps, and good luck!