Question #ba263

If you take any function such that:

#lim_(x->0) f(x) = 1#

and you multiply by #a# then:

#lim_(x->0) af(x) = a#

For instance:

#lim_(x->0) x^x#

is in the form #0^0#. We can evaluate the limit by writing the function as:

#x^x = e^(xlnx)#

Now:

#lim_(x->0) xlnx = lim_(x->0^+) lnx /(1/x)#

which is in the form #(-oo)/(+oo)# and can be evaluated using l'Hospital's rule:

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

and as #e^x# is a continuous function:

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

But then:

#lim_(x->0) 2*x^x = 2#

The example above is mentioned in Rotando & Korn "The Indeterminate Form #0^0#". (Mathematics Magazine,Vol. 50, No. 1 (January 1977), pp. 41-42.)

Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0, then f(x)^(g(x)) approaches 1 as x approaches 0 from the right.
I found this reference online at https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html

They show that being infinitely differentiable is not sufficient by counterexample:

#f(x) = {(e^(-1/x^2),"if",x != 0),(0,"if",x = 0):}#.

Then #lim_(xrarr0^+)(f(x))^x = 0#

while #lim_(xrarr0^-)(f(x))^x = oo#.