这些题的极限怎么求?
1个回答
展开全部
(1)原式=lim(x->0) (3x)/(5x)
=3/5
(2)原式=lim(x->0) x/tanx
=1
(3)原式=lim(x->0) (1-cosx)/(cosx*x^2)
=lim(x->0) [(1/2)*x^2]/(x^2)
=1/2
(4)原式=lim(x->0) x*sin(1/x)
=无穷小量*有界量
=0
(5)原式=lim(x->∞) [(1-1/x)^(-x)]^(-2)
=e^(-2)
(6)原式=lim(x->0) {(1+3tanx)^[1/(3tanx)]}^3
=e^3
(7)原式=lim(x->∞) {[1+1/(x+2)]^(x+2)}^[(x+3)/(x+2)]
=lim(x->∞) e^[(1+3/x)/(1+2/x)]
=e
(8)原式=lim(n->∞) {(1-2/n^2)^[-(n^2)/2]}^(-2/n)
=e^0
=1
(9)原式=lim(x->∞) [(3x^2+5)/(5x+3)]*(2/x)
=lim(x->∞) (6x^2+10)/(5x^2+3x)
=lim(x->∞) (6+10/x^2)/(5+3/x)
=6/5
=3/5
(2)原式=lim(x->0) x/tanx
=1
(3)原式=lim(x->0) (1-cosx)/(cosx*x^2)
=lim(x->0) [(1/2)*x^2]/(x^2)
=1/2
(4)原式=lim(x->0) x*sin(1/x)
=无穷小量*有界量
=0
(5)原式=lim(x->∞) [(1-1/x)^(-x)]^(-2)
=e^(-2)
(6)原式=lim(x->0) {(1+3tanx)^[1/(3tanx)]}^3
=e^3
(7)原式=lim(x->∞) {[1+1/(x+2)]^(x+2)}^[(x+3)/(x+2)]
=lim(x->∞) e^[(1+3/x)/(1+2/x)]
=e
(8)原式=lim(n->∞) {(1-2/n^2)^[-(n^2)/2]}^(-2/n)
=e^0
=1
(9)原式=lim(x->∞) [(3x^2+5)/(5x+3)]*(2/x)
=lim(x->∞) (6x^2+10)/(5x^2+3x)
=lim(x->∞) (6+10/x^2)/(5+3/x)
=6/5
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