Question

用洛必达法则求极限 lim→正无穷x×[(根号x^2+1)-x]

Answer 1

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