微积分问题：6道计算题，写出过程，如图所示。

Question

如题所述

Answer 1

1.原式=lim(x→1)(x^3-x^2+x^2-x-2x+2)/(x^3-x^2-x+1)=lim(x→1)(x^2(x-1)+x(x-1)-2(x-1))/(x^2(x-1)-(x-1))=lim(x→1)(x^2+x-2)/(x^2-1)=lim(x→1)[(x+2)(x-1)]/[(x+1)(x-1)]=lim(x→1)(x+2)/(x+1)=3/2

2.两边对x求导：y+xy'=e^(x+y)*(1+y')
y+xy'=e^(x+y)+y'e^(x+y)
y'(x-e^(x+y))=e^(x+y)-y
y'=(e^(x+y)-y)/(x-e^(x+y))

3.因为0<=|y|=|x^2sin(1/x)|<=|x^2|
由夹逼定理知lim(x→0)y=0=y(0)
而x≠0时y为初等函数，所以y连续
lim(x→0)(y-y(0))/(x-0)=lim(x→0)(x^2sin(1/x)-0)/x=lim(x→0)xsin(1/x)
因为0<=|xsin(1/x)|<=|x|
由夹逼定理知lim(x→0)xsin(1/x)=0，即y'(0)=0
而x≠0时y'=...为初等函数，所以y可导

4.原式=lim(x→∞)(1-2/(x+1))^x=lim(x→∞)[(1-2/(x+1))^(-(x+1)/2)]^(-2)=e^(-2)=1/e^2

5.y'=cosxln(x^2+x)+sinx*1/(x^2+x)*(2x+1)=cosxln(x^2+x)+sinx*(2x+1)/(x^2+x)

6.两边对x求导：y'=e^y+xy'e^y
y'=e^y/(1-xe^y)
即dy=e^y/(1-xe^y)*dx