Question

How do you determine the limit of `(8/x^3) + 8` as x approaches 0+?

Answer 1

`lim_(x->0^+) (8/x^3)+8 =+oo`

Explanation:

When `x` approaches `0` from positive values, it is easy to see that `(8/x^3)` is positive and not bounded so that:

`lim_(x->0^+) (8/x^3)+8 =+oo`

More formally, given `M>0` let's take `delta < 1/M^(1/3)`.

For `x in (0,delta)`:

`|8/x^3+8| > |8/x^3| > |8/delta^3| > |8/(1/M^(1/3))^3|>|8M| > M`

which proves the point.