求极限lim(x→0)][ln(1+2x^2)]/(1-cosx),急需要作答,

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求极限lim(x→0)][ln(1+2x^2)]/(1-cosx),急需要作答,

求极限lim(x→0)][ln(1+2x^2)]/(1-cosx),急需要作答,
求极限lim(x→0)][ln(1+2x^2)]/(1-cosx),急需要作答,

求极限lim(x→0)][ln(1+2x^2)]/(1-cosx),急需要作答,
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

原式=lim(x→0)2x^2/(2sin^2(x/2)) (ln(1+x)~x)
=lim(x→0)x^2/(x/2)^2 (sinx~x)
=4