这是一个0/0型的极限,可以采用洛必达法则.
lim【x→0】[ln(1+2x²)]/(1-cosx)
lim【x→0】[ln(1+2x²)]'/(1-cosx)'
=lim【x→0】[4x/(1+2x²)]/(sinx)
=lim【x→0】[4x/(sinx+2x²sinx)]
=lim【x→0】[(4x)'/(sinx+2x²sinx)']
=lim【x→0】[4/(cosx+4xsinx+2x²cosx)]
=4/(1+0+0)
=4