Question

Question #ba263

Answer 1

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

Explanation:

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`

Answer 2

`lim_(xrarr0^+) x^(a/lnx) = e^a` (Although this is a function that simplifies to a constant function.)

Explanation:

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`.