limx→0 (e^x-cosx-2x)/x^2-2x 几道AP微积分题目求教!1 求题目中这个极限的值2 如果g(x)=∫(上2x下0)f(t)dt 求g"(3)的值?3 原题是the number of moose in a national park is modeled by the function M that satisfies the logistic di

Question

limx→0 (e^x-cosx-2x)/x^2-2x 几道AP微积分题目求教!
1 求题目中这个极限的值
2 如果g(x)=∫(上2x下0)f(t)dt 求g'(3)的值?
3 原题是
the number of moose in a national park is modeled by the function M that satisfies the logistic differential equation dM/dt =0.6M(1-M/200),where t is the time in years and M(0)=50,what is lim(t→无穷)M(t)?
4 无穷级数∑(n=1到无穷) n/(n^p +1)收敛 求p取值范围
5 ∫2x/(x+2)(x+1) dx=?我感觉这个题应该用分部积分求解，但是总是就被绕进去了= π

Answer 1

1,罗比达法则,上下求导后
limx→0 (e^x+sinx-2) / (2x-2)
分式上下分别趋向 -1和-2 是定值,所以答案是1/2
2.g'(x)=2f(2x),g'(3)=2f(6)
3.解微分方程200dM/[M(200-M)]=(0.6)dt
[(1/M)+1/(200-M)]dM=0.6dt
ln[M/(200-M)]=0.6t+C1
M/(200-M)=Ce^(0.6t),M(0)=50,C=1/3,
解出来M=200(e^(3t/5))/(3+e^(3t/5)),
求极限limM=200
4.un=n/(n^p+1)≤n/(n^p)=1/[n^(p-1)]
由题意,un收敛,1/[n^(p-1)]也收敛,
所以,p-1>1,
所以p>2
5.∫2x/(x+2)(x+1) dx= 2∫[2/(x+2)-1/(x+1)]dx,会了么?
=2[2ln(x+2)-ln(x+1)]+C
=2ln(x+2)²/(x+1)+C