How to find LIM (x → 0 +) [(x ^ x-1) / xlnx]?

How to find LIM (x → 0 +) [(x ^ x-1) / xlnx]?

lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(x→0+)[xlnx/xlnx]
=1
E ^ X-1 and X are equivalent infinitesimals

When the limit x tends to 0: Lim xlnx / (1-cosx) =?
I think the denominator 1-cosx ~ &# 189; X & # 178; becomes 2limlnx / X. isn't limlnx / x = ∞? 2 times ∞?

lim xlnx/(1-cosx)
=lim xlnx/(1/2x^2)
=lim2lnx/x=∞
The limit of this question is infinite, and the answer is wrong

The limit problem has zero for year n = 0,1,2

X (n + 1) - 1 = - (xn) & sup2; + 2xn-1 = - (XN-1) & sup2;, so the general term formula of sequence {XN-1} is (xn) - 1 = - (x (n-1) - 1) & sup2; = - (x (n-2) - 1) ^ 4 =-(x0-1) ^ (2n) from this we can get: xn = 1 - (x0-1) ^ (2n) LIM (n tends to infinity) xn = Lim [1 - (x0-1) ^ (2n)] = 1-lim (x0-1) ^ (2n)