Ta có: [TEX]\underset{x \to 0^+}{\lim} (x^x-1).\ln x=\underset{x \to 0^+}{\lim} ([/TEX][tex]\underset{x \to 0^+}{\lim} (e^{\ln (x^x)}-1)\ln x=\underset{x \to 0^+}{\lim} \frac{e^{x \ln x}-1}{x\ln x}.\frac{\ln ^2x}{\frac{1}{x}}[/tex]
Ta thấy: [TEX]\underset{x \to 0^+}{\lim} x\ln x =\underset{x \to 0^+}{\lim} \frac{\ln x}{\frac{1}{x}}=\underset{x \to 0^+}{\lim}\frac{(\ln x)'}{(\frac{1}{x})'}=\underset{x \to 0^+}{\lim} -x=0[/TEX]
Từ đó [TEX]\underset{x \to 0^+}{\lim} \frac{e^{x\ln x}-1}{x\ln x}=1[/TEX]
Lại có: [TEX]\underset{x \to 0^+}{\lim} \frac{\ln ^2(x)}{\frac{1}{x}}=\underset{x \to 0^+}{\lim} \frac{(\ln ^2x)'}{(\frac{1}{x})'}=\underset{x \to 0^+}{\lim} -2x\ln x=0[/TEX]
Suy ra [TEX]\underset{x \to 0^+}{\lim} (x^x-1).\ln x=0.1=0[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
