Question

What is `lim_(x->0) x/abs(x)` ?

Answer 1

This function has one-sided limits which disagree and therefore there is no two-sided limit.

Explanation:

The graph is of a function like:

`f(x) = { (1 " if " x <= 0), (-1 " if " x > 0) :}`

It has one-sided limits as `x` approaches zero, namely:

`lim_(x->0^-) f(x) = 1`

`lim_(x->0^+) f(x) = -1`

As these one-sided limits disagree, there is no value we can assign to `f(0)` which can make `f(x)` continuous there.

For example, for any `delta > 0`, we find:

`abs(f(delta/2) - f(-delta/2)) = abs(-1-1) = 2`

Hence for any `epsilon in (0, 2)` we cannot find a `delta > 0` such that `AA x_1, x_2 in (-delta, delta)` we have `abs(f(x_1)-f(x_2)) < epsilon`. Since this Cauchy condition fails, the function `f(x)` does not converge to a limit as `x->0`.