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你得告诉大家x→?啊,否则怎么求极限,还得要我们猜,我猜是x→0吧,如果不是,你再追问。
lim[x→0] [3x(2x+1)]/[(-x-1+√(1+x²)]
分母有理化
=lim[x→0] [3x(2x+1)][(x+1+√(1+x²)] / [(-x-1+√(1+x²)][(x+1+√(1+x²)]
=lim[x→0] [3x(2x+1)][(x+1+√(1+x²)] / [x²+1-(x+1)²]
=lim[x→0] [3x(2x+1)][(x+1+√(1+x²)] / (-2x)
=lim[x→0] [3(2x+1)][(x+1+√(1+x²)] / (-2)
=-3
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮。
lim[x→0] [3x(2x+1)]/[(-x-1+√(1+x²)]
分母有理化
=lim[x→0] [3x(2x+1)][(x+1+√(1+x²)] / [(-x-1+√(1+x²)][(x+1+√(1+x²)]
=lim[x→0] [3x(2x+1)][(x+1+√(1+x²)] / [x²+1-(x+1)²]
=lim[x→0] [3x(2x+1)][(x+1+√(1+x²)] / (-2x)
=lim[x→0] [3(2x+1)][(x+1+√(1+x²)] / (-2)
=-3
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮。
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