Question

Can someone please help me how do i solve the indeterminate limx->o^+ x^x^2=

Answer 1

Mmm....I suppose ln could do the trick. If we treat this as a situation where this is simply limx->0(+) [y = x^(x^2)], we could ln both sides to remove that x^2 exponent.
\[\lim_{x \rightarrow 0^{+}} lny = x^{2}lnx\] Using ln's properties to pull that x^2 down. Now we just need to turn this into an indeterminant form. One way to do this is to rewrite x^2 into 1/x^-2 Rewriting it like that gives me
\[\lim_{x \rightarrow 0^{+}}lny =\frac{lnx}{x^{-2}}\]Now this can be considered in undeterminant form when 0 is plugged in, meaning we can do l'hopitals rule. So we would do the derivative of the numerator and the derivative of the denominator separately.
derivative of lnx is 1/x
derivative of x^-2 is -2x^-3, leaving us now with:
\[\lim_{x \rightarrow 0^{+}}lny =\frac{ 1 }{ x*-2x^{-3} } \implies -\frac{ x^{2} }{ 2 }\]
Taking the limit to 0 leaves you left with ln(y) = 0. But then you have to solve for y, so taking e to both sides you get y = e^0, which is 1.