當x→0，求lim (√x．sinx)/(x^(2/3))？
回答 (1)
2018-01-30 7:35 am
Sol
lim(x->0)_(√x－Sinx)/[x^(2/3)]
Set x=y^6
√x=y^3
x^(2/3)=y^4
lim(x->0)_(√x－Sinx)/[x^(2/3)]
=lim(y->0)_[y^3－Sin(y^6)]/y^4 0/0 type
=lim(y->0)_[3y^2－6y^5Cos(y^6)]/(4y^3)
=lim(y->0)_[3－6y^3Cos(y^6)]/(4y)
So
lim(x->0+)_(√x－Sinx)/[x^(2/3)]
=∞
lim(x->0-)_(√x－Sinx)/[x^(2/3)]
=－∞
lim(x->0)_(√x－Sinx)/[x^(2/3)]
Set x=y^6
√x=y^3
x^(2/3)=y^4
lim(x->0)_(√x－Sinx)/[x^(2/3)]
=lim(y->0)_[y^3－Sin(y^6)]/y^4 0/0 type
=lim(y->0)_[3y^2－6y^5Cos(y^6)]/(4y^3)
=lim(y->0)_[3－6y^3Cos(y^6)]/(4y)
So
lim(x->0+)_(√x－Sinx)/[x^(2/3)]
=∞
lim(x->0-)_(√x－Sinx)/[x^(2/3)]
=－∞
收錄日期: 2021-04-30 22:39:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180129131338AAaGgKn