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

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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

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