Question

How do you evaluate `[ ( x )^(2x) ]` as x approaches 0+?

Answer 1

I don't think this can be done algebraically. I think you'll need l"Hopital's rule.

Explanation:

`x^(2x) = e^(2xlnx)`

Since the exponential function is continuous, we can simplify the problem to find `lim_(xrarr0^+)xlnx = L`. (The limit we want will then be `e^(2L)`.)

`lim_(xrarr0^+)xlnx` has initial form `0*(-oo)`. This form is indeterminate.

We can apply l'Hopital''s Rule if we can make the form `0/0` or `+-oo/oo`.

`lim_(xrarr0^+)xlnx = lim_(xrarr0^+) lnx/(1/x)` has form `-oo/oo`

Applying the rule gets us

`= lim_(xrarr0^+) (1/x)/(-1/x^2) = lim_(xrarr0^+)( -x) = 0`.

Therefore, `lim_(xrarr0^+)(2xlnx) = 0`

We conclude that

`lim_(xrarr0^+) x^(2x) = lim_(xrarr0^+) e^(2xlnx)`

`= e^((lim_(xrarr0^+) 2xlnx))`

`= e^0 = 1`