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

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

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