用洛必达法则求极限 lim→正无穷x×[(根号x^2+1)-x]
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/09 09:49:50
用洛必达法则求极限 lim→正无穷x×[(根号x^2+1)-x]
没有用洛必达法则:
lim(x→∞) x[√(x²+1)-x]
=lim(x→∞) x[√(x²+1)-x][√(x²+1)+x]/[√(x²+1)+x],分子有理化
=lim(x→∞) x(x²+1-x²)/[√(x²+1)+x]
=lim(x→∞) x/[√(x²+1)+x],若需要,这步可以用洛必达法则上下求导...①
=lim(x→∞) 1/[√(x²+1)/x+1],上下除x
=lim(x→∞) 1/{√[(x²+1)/x²]+1}
=lim(x→∞) 1/[√(1+1/x²)+1]
=1/[√(1+0)+1]
=1/(1+1)
=1/2
用洛必达法则:由①
=lim(x→∞) x/[√(x²+1)+x]
=lim(x→∞) 1/[1/2√(x²+1)*2x+1],上下求导
=lim(x→∞) 1/[x/√(x²+1)+1]
=lim(x→∞) √(x²+1)/[x+√(x²+1)],分母进行通分
=lim(x→∞) [√(x²+1)/x]/[1+√(x²+1)/x],上下除x
=lim(x→∞) √(1+1/x²)/[1+√(1+1/x²)]
=√(1+0)/[1+√(1+0)]
=1/(1+1)
=1/2
lim(x→∞) x[√(x²+1)-x]
=lim(x→∞) x[√(x²+1)-x][√(x²+1)+x]/[√(x²+1)+x],分子有理化
=lim(x→∞) x(x²+1-x²)/[√(x²+1)+x]
=lim(x→∞) x/[√(x²+1)+x],若需要,这步可以用洛必达法则上下求导...①
=lim(x→∞) 1/[√(x²+1)/x+1],上下除x
=lim(x→∞) 1/{√[(x²+1)/x²]+1}
=lim(x→∞) 1/[√(1+1/x²)+1]
=1/[√(1+0)+1]
=1/(1+1)
=1/2
用洛必达法则:由①
=lim(x→∞) x/[√(x²+1)+x]
=lim(x→∞) 1/[1/2√(x²+1)*2x+1],上下求导
=lim(x→∞) 1/[x/√(x²+1)+1]
=lim(x→∞) √(x²+1)/[x+√(x²+1)],分母进行通分
=lim(x→∞) [√(x²+1)/x]/[1+√(x²+1)/x],上下除x
=lim(x→∞) √(1+1/x²)/[1+√(1+1/x²)]
=√(1+0)/[1+√(1+0)]
=1/(1+1)
=1/2
用洛必达法则求极限 lim→正无穷x×[(根号x^2+1)-x]
求极限.lim x(根号下(x^2+1) ) -x x趋向正无穷
求lim(x-正无穷)2x^2-3x-4/根号下x^4+1的极限
【求极限】(x趋向正无穷)lim((x^2)/arctanx).
ln(1+2x^2)/ln(1+3x^3) lim趋向于正无穷,用洛必达法则求它的极限
用洛必达法则求极限,x趋近正无穷,(x^3+x^2+x+1)的三次方根—x.
2.(1)求极限lim (x→0) sin2x /根号x+1 -1 (2)求极限lim (x→无穷) lim (x+2
求极限lim(根号(x+根号(x+根号x ))-根号x),(x趋向于正无穷)
lim(根号下(x^2+x+1)减根号下(x^2-x+1))x趋向于正无穷求极限详细过程
lim(根号下(x^2+x+1)-根号下(x^2-x+1))x趋向于正无穷求极限详细过程
求lim(x→正无穷){√(x^2+4)-2/x}的极限
洛必达法则求极限 lim(x-0) ln(x+根号(1+x^2))