You need to use l'Hospital's theorem such that:

`lim_(x->x_0) f^g = lim_(x->x_0) e^(ln(f^g))`

Using the relation above, yields:

`lim_(x->0,x>0) x^x = lim_(x->0,x>0) e^(ln(x^x))`

Using logarithmic power identity yields:

`lim_(x->0,x>0) e^(ln(x^x)) = lim_(x->0,x>0) e^(x ln x)`

`lim_(x->0,x>0) e^(x ln x) = e^(lim_(x->0,x>0) x ln x)`

You need to evaluate the limit of exponent such that:

`lim_(x->0,x>0) x ln x = lim_(x->0,x>0) (ln x)/(1/x) = oo/oo`

The indetermination oo/oo requests for you to use l'Hospital's theorem again yields:

`lim_(x->0,x>0) (ln x)/(1/x) = lim_(x->0,x>0) ((ln x)')/((1/x)')`

`lim_(x->0,x>0) ((ln x)')/((1/x)') = lim_(x->0,x>0) (1/x)/(-1/x^2) = lim_(x->0,x>0) -x = 0`

Hence, evaluating the limit yields:

`lim_(x->0,x>0) e^(x ln x) = e^0 = 1`

**Hence, evaluating the limit using l'Hospital's theorem, yields `lim_(x->0,x>0) x^x = 1` .**