How do you find the limit of #x^(2x)# as x approaches 0?

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Answer 1 · 2016-06-22T07:58:04

#1#

#L = lim_{x \to 0} x^{2x}#

consider
#y = x^{2x}# so that # ln y = 2x ln x#

# ln L = lim_{x \to 0} 2x ln x#

# = 2 lim_{x \to 0} (ln x)/(1/x)# so it is now indeterminate

Applying L'Hopital's Rule

#ln L = 2 lim_{x \to 0} (ln x)/(1/x) = 2 lim_{x \to 0} (1/x)/(-1/x^2) = -2 lim_{x \to 0} x = 0#

so # lim_{x \to 0} ln L = 0 \implies lim_{x \to 0} L = 1#