问题及解答

求极限 $\underset{x\to 0^{+}}{lim}(x^x-1)\ln x$.

注意到 $x^x-1=e^{x\ln x}-1$, 而

$$\underset{x\to 0^{+}}{lim}x\ln x\stackrel{t=1/x}{=}\underset{t\to +\mathrm{\infty }}{lim}\frac{1}{t}\ln \frac{1}{t}=-\underset{t\to +\mathrm{\infty }}{lim}\frac{\ln t}{t}=0.$$

因此 $e^{x\ln x}-1\sim x\ln x$, 当 $x\to 0^{+}$ 时. 于是

$$\begin{array}{rl}\underset{x\to 0^{+}}{lim}(x^x-1)\ln x& =\underset{x\to 0^{+}}{lim}(e^{x\ln x}-1)\ln x\\ & =\underset{x\to 0^{+}}{lim}(x\ln x)\cdot \ln x\\ & =\underset{x\to 0^{+}}{lim}x{\ln }^{2}x\\ & \stackrel{x=1/t}{=}\underset{t\to +\mathrm{\infty }}{lim}\frac{{\ln }^2(\frac{1}{t})}{t}\\ & =\underset{t\to +\mathrm{\infty }}{lim}\frac{(\ln t{)}^2}{t}\\ & =\underset{t\to +\mathrm{\infty }}{lim}(\frac{\ln t}{\sqrt{t}}{)}^2\\ & =\underset{t\to +\mathrm{\infty }}{lim}4\cdot (\frac{\ln \sqrt{t}}{\sqrt{t}}{)}^2\\ & =0.\end{array}$$