Using l'Hopital' Rule, find the limit of x^x^2 as x approaches 0 from the right.

We have to find the value of lim x-->0+
[(x^x)^2)]

lim x-->0+
[(x^x)^2)]

=> lim x-->0+ [e^(ln
((x^x)^2))]

=> lim x-->0+ [e^(ln
((x^2x))]

=> lim x-->0+ [e^(2x*ln
x)]

As the exponential function is continuous we can write
this as

=> e^[lim x-->0+ (2x*ln
x)]

As 2x*ln is of the indeterminate form we can use
l'Hopital's rule

lim x-->0+ (2x*ln
x)

=> lim x-->0+ (2*ln
x/(1/x))

=> lim x-->0+
(2*(1/x)/(-1/x^2)

=> lim x-->0+
(-2*x^2/x)

=> lim x-->0+
(-2*x)

substitute x =
0

=> 0

Now we have e^0
= 1

This gives the value of lim x-->0+
[(x^x)^2)] = 1