二分之π减arctanx

从广质4844求limx→∞(π/2 - arctanx)/0.5x??? -
娄枫罗15393978626 ______ X=π/2-arctanx x→∞,π/2-arctanx→0,X→0 limx→∞(π/2-arctanx)/0.5x =0

从广质4844高数极限证明lim(x>无穷)arctanx=二分之派 -
娄枫罗15393978626 ______[答案] 应该是:x→+∞ |arctanx-π/2|=π/2-arctanx 对于任意的正数ε(ε<1),要使得|arctanx-π/2|<ε,只要π/2-arctanx<ε,即x>cotε即可. 取正数X=cotε,当x>X时,|arctanx-π/2|<ε 所以,lim(x→+∞) arctanx=π/2

从广质4844lim(x→∞)x(π/2 - arctanx) -
娄枫罗15393978626 ______ 不用洛必达法则 设arctanx=t.x=tant lim(x→∞)x(π/2-arctanx) =lim(t→π/2)tant(π/2-t) =lim(t→π/2)sint*[(π/2-t)/sin(π/2-t)] =1 ((π/2-t)/sin(π/2-t)趋于1,特殊极限)

从广质4844lim(x→∞)x(π/2 - arctanx)求过程(最好不用洛必达法则) -
娄枫罗15393978626 ______[答案] 不用洛必达法则 设arctanx=t.x=tant lim(x→∞)x(π/2-arctanx) =lim(t→π/2)tant(π/2-t) =lim(t→π/2)sint*[(π/2-t)/sin(π/2-t)] =1 ((π/2-t)/sin(π/2-t)趋于1,特殊极限)

从广质4844求1/(π/2 - arctanx)的导数 -
娄枫罗15393978626 ______[答案] 1/(π/2-arctanx)=-(π/2-arctanx)`/(π/2-arctanx)²=1/[(1+x²)((π/2-arctanx)²] 利用商的求导公式(u/v)`=(u`v-uv`)/v²

从广质4844高等数学,怎么样用极限的定义证明arctanx在正无穷大处趋向于二分之π -
娄枫罗15393978626 ______ 根据极限定义,|arctanx-π/2|=|arccotx|,对于任意的ε>0,存在N=[cotε]+1,使得当n>N时,有[arctanx-π/2]

从广质4844lim(x→+∞)x(π/2 - arctanx)求详解 -
娄枫罗15393978626 ______[答案] x(π/2-arctanx)=(π/2-arctanx)/(1/x)是0/0型,用洛必达法则,上下求导得(1/(1+x^2))/(1/x^2) =x^2/(1+x^2)=1-1/(1+x^2)=1

从广质4844lim x(π/2 - arctanx)=?lim下面是+∞ -
娄枫罗15393978626 ______[答案] 方法1lim[x(π/2-arctanx)]=lim(π/2-arctanx)/(1/x) (0/0型,用洛必达法则)=lim(1/(1+x^2))/(1/x^2)=limx^2/(1+x^2)=lim[1-1/(1+x^2)]=1方法2设arctanx=t.则x=tantlim(x→∞)x(π/2-arctanx)=lim(t→π/2)ta...

从广质4844y=x(π/2 - arctanx)的水平渐近线怎么求 -
娄枫罗15393978626 ______[答案] y=x(π/2-arctanx) 所以 lim(x->∞)y=lim(x->∞)x(π/2-arctanx) =lim(x->∞)(π/2-arctanx)/(1/x) =lim(x->∞)(-1/(1+x²))/(-1/x²) =lim(x->∞)x²/(1+x²) =lim(x->∞)2x/2x =1 所以 水平渐近线为y=1

从广质4844求x(二分之派 - arctanX)当x趋近正无穷的极限~ -
娄枫罗15393978626 ______[答案] -pi*pi/4