求arctanx除以x的极限

阙翰瑶1236对arctanx除x²(1+x²)求不定积分 -
颜鸦例18648625473 ______[答案] 把分母裂开成两项相减: ∫ dx \\/x²(1+x²)=∫ [(1\\/x²)-1\\/(1+x²)] dx=-(1\\/x)+arctanx+C

阙翰瑶1236证明当x趋近于0时arccotx除以x的极限等于1 -
颜鸦例18648625473 ______ 错了,应该是 lim(x→0)(arctanx/x) = lim(x→0)(t/tant) (x = tant) = lim(x→0)(t/sint)*cost = 1*1 = 1.

阙翰瑶1236arctanx÷y的原函数 -
颜鸦例18648625473 ______ ∫zhidao arctanx dx= x * arctanx - ∫ x d(arctanx)= x * arctanx - ∫ x/(1+x²) dx= x * arctanx - (1/2)∫ d(x²)/(1+x²)= x * arctanx - (1/2)∫ d(1+x²)/(1+x²)= x * arctanx - (1/2)ln(1+x²) + C

阙翰瑶1236证明当x趋近于0时arccotx除以x的极限等于1 -
颜鸦例18648625473 ______[答案] 错了,应该是 lim(x→0)(arctanx/x) = lim(x→0)(t/tant) (x = tant) = lim(x→0)(t/sint)*cost = 1*1 = 1.

阙翰瑶1236求极限 f(x)=arctanx/x -
颜鸦例18648625473 ______ 上下分别求导,arctanx求导=1/(1+x²),分母求导为1, 所以f(x)=arctanx/x的极限就等于1/(1+x²)的极限, 当x趋于无穷大时 1/(1+x²)趋于0

阙翰瑶1236lim(x+arctanx)/(x - arctanx) (x→∞) -
颜鸦例18648625473 ______ arctanx在x →∞的时候趋向于pi/2 所以arctanx/x = 0 上下同时除以x得 (1+arctanx/x)/(1-arctanx/x)=1/1=1

阙翰瑶1236f(x)的原函数是(arctanx)/x 求f(x) -
颜鸦例18648625473 ______ f(x)的原函数是(arctanx)/x f(x)=((arctanx)/x )'=(x/(x^2+1)-arctanx)/x^2=1/(x(x^2+1))-1/x^2 *arctanx

阙翰瑶1236求极限解题lim(x→∞)=arctanx/x -
颜鸦例18648625473 ______[答案] 由罗比塔法则得 lim(x→∞)arctanx/x =lim(x→∞)1/(x2+1)=0

阙翰瑶1236求解lim(x - >+ - 无穷)arctanx - x的解的过程, -
颜鸦例18648625473 ______[答案] 问题有点奇怪,你看看有没抄错的地方.不过可以求 首先有 lim(x->+无穷)arctanx=π/2 lim(x->-无穷)arctanx=-π/2 因此 lim(x->+无穷)arctanx-x=-无穷 lim(x->-无穷)arctanx-x=+无穷 所以极限是不存在的.

阙翰瑶1236求y=arctanx+arctan(1 - x)/(1+x)的值 -
颜鸦例18648625473 ______[答案] 求y=arctanx+arctan[(1-x)/(1+x)]的值 tany=[tan(arctanx)+tanarctan(1-x)/(1+x)]/[1-tan(arctanx)tanarctan(1-x)/(1+x)] =[x+(1-x)/(1+x)]/[1-x(1-x)/(1+x)]=[x(1+x)+(1-x)]/[(1+x)-x(1-x)]=(x²+1)/(1+x²)=1 故y=π/4.