Question

How to solve that Limit without graphing ? `limx->0^+` `1/x` = `oo` `limx->0^-` `1/x` `-oo`

Answer 1

Plug in a number for `x` such as `0.0000001` which is `0^+`

Explanation:

Given: `lim x-> 0^+ 1/x = oo; " "lim x-> 0^- 1/x = -oo`

Without graphing you can plug in a value that approaches `0^+` and `0^-`.

`x->0^+` means approach from the right side (positive) of zero:
Let `x = .01; " " 1/x = 100`

Let `x = .0001; " " 1/x = 10,000`

Let `x = .000001; " " 1/x = 1,000,000`

So, as `x ->0^+ 1/x = oo`

`x->0^-` means approach from the left side (negative) of zero:
Let `x = -.01; " " 1/x = -100`

Let `x = -.0001; " " 1/x = -10,000`

Let `x = -.000001; " " 1/x = -1,000,000`

So, as `x->0^-` `1/x = -oo`