What is #lim_(xrarr0+) ( x )^(2x)#?

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Answer 1 · 2015-11-02T18:10:09

#lim_(xrarr0+) ( x )^(2x) = 1#

#x^(2x)= e^ln(x^(2x)) = e^(2xlnx)#

We shall find #lim_(xrarr0^+)2xlnx = L#

So that #lim_(xrarr0^+)x^(2x) = e^L#

#lim_(xrarr0^+)2xlnx# has indeterminate form #0*-oo#, so we rewrite:

#lim_(xrarr0^+)2xlnx = 2lim_(xrarr0^+)lnx/(1/x)# which now has form #oo/oo#

So we can apply l'Hopital's Rule:

#2lim_(xrarr0^+)lnx/(1/x) = 2lim_(xrarr0^+)(1/x)/(-1/x^2)#

# = 2lim_(xrarr0^+)x = 0#

We conclude that

#lim_(xrarr0+) ( x )^(2x) = lim_(xrarr0+) e^(2xlnx) = e^0 = 1#