1.這個(gè)函數(shù)可以求導(dǎo),易知該函數(shù)的定義域?yàn)閄＞0
∵x=e^lnx
設(shè)f(x)=x^x=e^(xlnx)
f′(x)=e^(xlnx)·(xlnx)′
=e^(xlnx)·(1+lnx)=x^x(1+lnx)
令f′(x)＞0,解得x>1/e
f′(x)＜0,解得0這里要用到一個(gè)重要的函數(shù)極限，limx→﹢∞（1+1/x）^x=e[（1+1/x）^x]′=（1+1/x）^x[ln(1+1/x)﹣1/1+x]∵在﹙0，+∞﹚上，（1+1/x）^x＞0恒成立∴只需判斷l(xiāng)n(1+1/x)﹣1/1+x的符號(hào)對(duì)h(x)=ln(1+1/x)﹣1/1+x求導(dǎo)得，h′(x)＝-1/x(x+1)²＜0∴h(x)=ln(1+1/x)﹣1/﹙1+x﹚在﹙0，+∞﹚上單調(diào)遞減∵ln(1+1/x)﹣1/﹙1+x﹚=[（x+1）ln(1+1/x)﹣1]／(x+1)又∵（x+1）ln(1+1/x)＞xln(1+1/x)根據(jù)極限的保不等式性limx→﹢∞（x+1）ln(1+1/x)＞limx→﹢∞xln(1+1/x)=limx→﹢∞ln(1+1/x)^x=lne=1∴l(xiāng)imx→﹢∞（x+1）ln(1+1/x)＞1∴在﹙0，+∞﹚上（x+1）ln(1+1/x)﹣1＞0恒成立∴[（x+1）ln(1+1/x)﹣1]／（x+1）＞0即ln(1+1/x)﹣1/1+x＞0∴（1+1/x）^x[ln(1+1/x)﹣1/1+x]＞0∴﹙1+1/x﹚^x在﹙0,﹢∞﹚上單調(diào)遞增這是一個(gè)重要的極限，是前幾倍的數(shù)學(xué)家發(fā)現(xiàn)的