Question

Why is it impossible to have `lim_(x->0) f(x)` and `lim_(f(x)->0)f(x)` simultaneously exist for any of these graphs?

`A)` `f(x) = 1/x^2`
`B)` `f(x) = -1/x^2`
`C)` `f(x) = 1/x`
`D)` `f(x) = -1/x`

Answer 1

All four of these graphs have the `x`-axis as a horizontal asymptote as `x-> pm infty` and the `y`-axis as a vertical asymptote as `x->0` from the right or left.

Answer 2

Well, by definition, a vertical asymptote is when at `x -> a`, `y -> pmoo` from either side and `x` never touches `a`. Similarly, a horizontal asymptote is when at `x -> pmoo`, `y -> b` from either side without ever reaching `b`.

For the function

`y = c/x`,

where `c` is a constant, if `x->0`, `y -> pmoo` from either side of `x = 0`, so you have a vertical asymptote. You can also find that as `x -> pmoo`, `y -> 0` but doesn't get there.

But if you have `x -> 0` and consequently `y -> pmoo`, you can't also have `x -> pmoo` so that `y -> 0`. It's not possible to approach both asymptotes at once because `x` cannot approach `0` and `pmoo` at the same time, and `y` cannot approach `pmoo` and `0` at the same time.

(Imagine trying to run to two different places at once; can't do it.)

Both kinds of asymptotes are on the graph, to be sure, but you can only approach one of those kinds of asymptotes at a time.