设函数f(x)连续,求lim(x→b)x∧2/(x-b)∫(x,b)f(t)dt,求解
人气:132 ℃ 时间:2020-05-27 17:35:29
解答
(1)落毕达法则:上下同时求导,分子为x∧2*∫(x,b)f(t)dt,分母为(x-b),分母求导为1,分子求导为2x∫(x,b)f(t)dt-x^2*f(x),此时带入x=b,可以求得极限为-b^2f(b)(2)定义:根据导数的定义求,lim(x→b)g(x)/(x-b)相当于...
推荐
- 设函数f(x)在区间[a,b]上连续,则lim(x->a)∫(a->x)f(t)dt=____,lim(x->a)1/(x-a)∫(a->x)f(t)dt=_____
- 设函数f(x)在[A,B]上连续,证明lim(h→0) 1/h*∫(x,a)[f(t+h)-f(t)]dt=f(x)-f(a),其中A
- 证明:若函数f(x)∈C[0,+∞],且lim(x->+∞)f(x)=A,则lim(x->+∞)[1/x*∫(0->x)f(t)dt]=A
- 设连续函数f﹙x﹚满足lim﹙x→0﹚f﹙x﹚/x=2 ,令F﹙x﹚=∫﹙0,1﹚f﹙xt﹚dt ,则F′﹙0﹚= _______
- 设f(x)是连续函数,F(x)=1/(x-a)∫[a,x]f(t)dt,则lim(x→a)F(x)=
- (2005x)的平方+2004×2006x=1,
- The computer revolution may well change society as ______ as did the Industrial Revolution.
- 如图,四边形EFGH为空间四边形ABCD的一个截面,若截面为平行四边形,求证:BD∥面EFGH.
猜你喜欢