证明:当x>1x > 1x>1时,lnx>2(x−1)x+1\ln x>\dfrac{2(x - 1)}{x + 1}lnx>x+12(x−1).
若函数在一点处极限存在,则在该点处一定连续.( )
已知y=f(3−x)y = f(3 - x)y=f(3−x),f(x)f(x)f(x)可导,则dy=f′(3−x)dxdy = f'(3 - x)dxdy=f′(3−x)dx.( )
Copyright ©