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

1 Answer
Jim H
Oct 31, 2016

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#

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