Question

求解高数新大一极限题

四道都要 谢谢 不要用什么洛必达法则 我们还没教过 我看不懂😭

Answer 1

如图所示

追问

第一题答案写的是6/5哎

Answer 2

(3)
lim(x->∞) [(3x^2+5)/(5x+3)] .sin(2/x)
=lim(x->∞) [(3x^2+5)/(5x+3)] .(2/x)
分子分母同时除以 x^2
=lim(x->∞) [(3+5/x^2)/(5+3/x)] .(2)
=6/5
(4)
lim(x->0) (x-sinx)/(x+sinx)
=lim(x->0) (x-x +(1/6)x^3)/(x+x)
=0
(5)
let
y=π/2-x
lim(x->π/2) (1+cosx)^(2secx)
=lim(x->π/2) (1+cosx)^(2/cosx)
=lim(y->0) (1+siny)^(2/siny)
=lim(y->0) (1+y)^(2/y)
=e^2
(6)
let
1/y = 1/(x+1)
lim(x->∞) [ x/(x+1)]^(x+3)
=lim(x->∞) [ 1- 1/(x+1)]^(x+3)
=lim(y->∞) ( 1- 1/y)^(y-1+3)
=lim(y->∞) ( 1- 1/y)^(y+2)
=lim(y->∞) ( 1- 1/y)^y
=e^(-1)