Question

How do you determine the limit of `( (1 / x ) - (1/x^2) )` as x approaches 0+?

Answer 1

`lim_(x->0+) (1/x-1/x^2) = -oo`

Explanation:

When `x != 0` we have:

`1/x-1/x^2 = x/x^2-1/x^2 = (x-1)/x^2`

As `x->0+` the numerator tends to `0-1 = -1`, while the denominator tends to `0` through positive values.

So `lim_(x->0+) (1/x-1/x^2) = lim_(x->0+) ((x-1)/x^2) = -oo`