limx→0 (e^x-cosx-2x)/x^2-2x 几道AP微积分题目求教!
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
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
数学人气：775 ℃时间：2020-05-22 15:35:14
优质解答
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1/22.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=0.6dtln[M/(200-M)]=0....
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