\[\begin{aligned} \ln\,y ~&=~ \ln\,\left(\lim_{x \to 0+}~x^x\right)\\[4pt] &=~ \lim_{x \to 0+}~\ln\,x^x \quad\text{(pass the natural logarithm function inside the limit)}\\[4pt] &=~ \lim_{x \to 0+}~x\...lny=ln(limx→0+xx)=limx→0+lnxx(pass the natural logarithm function inside the limit)=limx→0+xlnx→0⋅(−∞)=limx→0+lnx1/x→−∞∞=limx→0+1/x−1/x2by L'H\^{o}pital's Rulelny=limx→0+(−x)=0Por lo tanto,\(\displaystyle\lim_{x \to …
