limx→∞ln(x+5)/√(x^2+1)(这是∞/∞型,运用洛必达法则得)
=limx→∞1/[(x+5)*2x/2√(x^2+1)]
=limx→∞√(x^2+1)/[(x+5)*x]=0
啊~题目是这样的~求解答~limx→∞ln(x+5/√x^2+1)
limx→∞ln(x+5/√x^2+1)=limx→∞ln√[(x^2+10x+25)/(x^2+1)]=limx→∞1/2ln[(x^2+10x+25)/(x^2+1)]=limx→∞1/2ln[1+(10x+24)/(x^2+1)]=limx→∞1/2ln(1+0)=0