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

Question

#lim_(x->0^+)##sqrt(x)^sqrt(x)#

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Answer 1 · 2017-12-07T02:51:44

I get #1#. Don't you just love when that happens?

graph{sqrtx^(sqrtx) [-2.923, 4.872, -1.046, 2.85]}

Anytime I see exponents I take the #ln#...

#ln L = ln lim_(x->0^+) sqrtx^(sqrtx)#

Since #sqrtx# is continuous for #x >= 0#, the #ln# of the limit is the limit of the #ln# of whatever's inside.

#=> ln L = lim_(x->0^+) ln(sqrtx^(sqrtx))#

#= lim_(x->0^+) sqrtxlnsqrtx#

Right now we have the form #0 cdot -oo#, so change to dividing by #1/sqrtx# to get:

#= lim_(x->0^+) lnsqrtx/(1/sqrtx)#

Now L'Hopital's rule can apply, since this is of the form #oo/oo#. Differentiate the numerator and denominator separately to get:

#= lim_(x->0^+) (1/sqrtx cdot 1/(2sqrtx))/(-1/2x^(-3//2))#

#= lim_(x->0^+) (1/(2x))/(-1/(2x^(3//2))#

#= lim_(x->0^+) -sqrt(2x)#

#= 0#

And remember, this was the #ln# of the limit, so the actual limit is

#e^(lnL) = color(blue)(L) = e^0 = color(blue)(1)#