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

1 Answer
Dec 14, 2017

#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]}