
求极限要过程
1个回答
展开全部
解:
令1/x=t,则x=1/t
x→∞,则t→0
lim [x²+6sin(1/x)]/(x³+2x+sinx)
x→∞
=lim [(1/t)²+6sint]/[(1/t)³+2(1/t)+sin(1/t)]
t→0
=lim [(1/t)²+6t]/[(1/t)³+2(1/t)+sin(1/t)]
t→0
=lim (t+6t⁴)/[1+2t²+t³·sin(1/t)]
t→0
=(0+0)/(1+0+0)
=0
令1/x=t,则x=1/t
x→∞,则t→0
lim [x²+6sin(1/x)]/(x³+2x+sinx)
x→∞
=lim [(1/t)²+6sint]/[(1/t)³+2(1/t)+sin(1/t)]
t→0
=lim [(1/t)²+6t]/[(1/t)³+2(1/t)+sin(1/t)]
t→0
=lim (t+6t⁴)/[1+2t²+t³·sin(1/t)]
t→0
=(0+0)/(1+0+0)
=0
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询