Question

请问这道极限题怎么算

请问这道极限题怎么算谢谢

Answer 1

n趋近于无穷大，则1/n趋近于0，故sin(1/n)近似等于1/n，sin(1/n+1)近似等于1/n+1，括号里就等于1/n(n+1)，题目就等于求1/(n+1)当n趋近于无穷大的极限，最后等于0。

Answer 2

f(x) = e^x - (1+ax)/(1+bx) 为x 的3价无穷小
x->0
e^x = 1+x+(1/2)x^2 + (1/6)x^3 +o(x^3)
(1+ax)/(1+bx)
=(1+ax)[1-bx + (bx)^2 -(bx)^3 +o(x^3) ]
= 1+(a-b)x +(-ab+b^2).x^2 + (-ab^2-b^3)x^3 +o(x^3)
e^x - (1+ax)/(1+bx)
=[1 -(a-b)] x + [ 1/2 - (-ab+b^2) ] x^2 + [ 1/6 -(-ab^2-b^3) ] x^3 +o(x^3)
f(x) = e^x - (1+ax)/(1+bx) 为x 的3价无穷小
=>
1-(a-b)= 0 (1)
1/2 - (-ab+b^2) =0 (2)
sub (1) into (2)
1/2 - [ -(1+b)b + b^2 ] = 0
1/2 +b = 0
b = -1/2
form (1)
1-(a-b)= 0
1-(a+1/2)= 0
a= 1/2
ie
(a,b) = (1/2 , -1/2)