Question

How do you find the limit of `x^x` as `x>0^-`?

Answer 1

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

Explanation:

Write the function as:

`x^x = (e^lnx)^x =e^(xlnx)`

Note now that:

`lim_(x->0^+) xlnx`

is in the indeterminate form `0*oo`. We can reconduce it to the form `oo/oo` and then apply l'Hospital's rule:

`lim_(x->0^+) xlnx = lim_(x->0^+) lnx/(1/x) = lim_(x->0^+) (d/dx lnx)/(d/dx 1/x) = lim_(x->0^+) (1/x)/(-1/x^2) = lim_(x->0^+) -x = 0`

As `e^x` is a continuous function near `0`:

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

graph{x^x [-10, 10, -5, 5]}