求极限 lim(x→+0)[ln(1/x)]^x

xfhaoym

luyuanhong

xfhaoym

谢谢陆老师。

王守恩

至少，电脑是这样出来的。
\(\displaystyle\lim_{x\to 0}\big(\ln\frac{1}{x}\big)^x=1\)
\(\displaystyle\lim_{x\to 1}\big(\ln\frac{1}{x}\big)^x=0\)
\(\displaystyle\lim_{x\to 2}\big(\ln\frac{1}{x}\big)^x=\ln(2)^2\)
\(\displaystyle\lim_{x\to 3}\big(\ln\frac{1}{x}\big)^x=-\ln(3)^3\)
\(\displaystyle\lim_{x\to 4}\big(\ln\frac{1}{x}\big)^x=\ln(4)^4\)
\(\displaystyle\lim_{x\to 5}\big(\ln\frac{1}{x}\big)^x=-\ln(5)^5\)
\(\displaystyle\lim_{x\to 6}\big(\ln\frac{1}{x}\big)^x=\ln(6)^6\)
\(\displaystyle\lim_{x\to 7}\big(\ln\frac{1}{x}\big)^x=-\ln(7)^7\)
\(\displaystyle\lim_{x\to 8}\big(\ln\frac{1}{x}\big)^x=\ln(8)^8\)
\(\displaystyle\lim_{x\to 9}\big(\ln\frac{1}{x}\big)^x=-\ln(9)^9\)

王守恩

电脑能出来，人脑能出来吗？
\(\displaystyle\lim_{x\to 0}\big(\ln\frac{x-1}{x+1}\big)^x=1\)
\(\displaystyle\lim_{x\to 2}\big(\ln\frac{x-1}{x+1}\big)^x=\ln(3)^2\)
\(\displaystyle\lim_{x\to 4}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{5}{3}\big)^4\)
\(\displaystyle\lim_{x\to 6}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{7}{5}\big)^6\)
\(\displaystyle\lim_{x\to 8}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{9}{7}\big)^8\)
\(\displaystyle\lim_{x\to 10}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{11}{9}\big)^{10}\)
\(\displaystyle\lim_{x\to 12}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{13}{11}\big)^{12}\)
\(\displaystyle\lim_{x\to 14}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{15}{13}\big)^{14}\)
\(\displaystyle\lim_{x\to 16}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{17}{15}\big)^{16}\)
\(\displaystyle\lim_{x\to 18}\big(\ln\frac{x-1}{x+1}\big)^x=\ln\big(\frac{19}{17}\big)^{18}\)

王守恩

人脑能出来吗？
人脑敢出来吗？！
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-0}\big)=\frac{1}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-1}\big)=\frac{2}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-2}\big)=\frac{3}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-3}\big)=\frac{4}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-4}\big)=\frac{5}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-5}\big)=\frac{6}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-6}\big)=\frac{7}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-7}\big)=\frac{8}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-8}\big)=\frac{9}{11}\)
\(\displaystyle\lim_{x\to\infty}\big(\frac{x}{11}\ln\frac{x+1}{x-9}\big)=\frac{10}{11}\)