高数的无穷小的比较.
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/13 09:32:01
高数的无穷小的比较.
![](http://img.wesiedu.com/upload/1/1d/11d4308ee5ca37387b03919bc1aaaf16.jpg)
![](http://img.wesiedu.com/upload/1/1d/11d4308ee5ca37387b03919bc1aaaf16.jpg)
![高数的无穷小的比较.](/uploads/image/z/18467072-8-2.jpg?t=%E9%AB%98%E6%95%B0%E7%9A%84%E6%97%A0%E7%A9%B7%E5%B0%8F%E7%9A%84%E6%AF%94%E8%BE%83.)
用等价无穷小替换可注意到
ln(1+t) t ,e^t - 1 t (t→0),
有
2)g.e.= lim(x→0)[(3sinx)/x]*lim(x→0)[1/(1+cosx)]
+ lim(x→0)[(x^2)cos(1/x)/x]*lim(x→0)[1/(1+cosx)]
= 3lim(x→0)(sinx/x)*lim(x→0)[1/(1+cosx)]
+ lim(x→0)[xcos(1/x)]*lim(x→0)[1/(1+cosx)]
= 3*(1/2) + 0*1
= 3/2.
3)g.e.= lim(x→0)(-sin2x)/(3x) = -2/3.
ln(1+t) t ,e^t - 1 t (t→0),
有
2)g.e.= lim(x→0)[(3sinx)/x]*lim(x→0)[1/(1+cosx)]
+ lim(x→0)[(x^2)cos(1/x)/x]*lim(x→0)[1/(1+cosx)]
= 3lim(x→0)(sinx/x)*lim(x→0)[1/(1+cosx)]
+ lim(x→0)[xcos(1/x)]*lim(x→0)[1/(1+cosx)]
= 3*(1/2) + 0*1
= 3/2.
3)g.e.= lim(x→0)(-sin2x)/(3x) = -2/3.