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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