如题所述
ln(1+x2)/(3x+1)是否是ln(1+x²)/(3x+1)
如果是
limln(1+x²)/(3x+1)=lim[ln(1/x+x)/(3+1/x)]x趋向于无穷大时为无穷大。
如果不是
limln(1+x2)/(3x+1)=limln[(1/x+2)/(3+1/x)]=ln(2/3)(x趋向于无穷大时)
如果是ln(1+x²)÷（3x+1）用罗必塔法则
lim[ln(1+x²)÷（3x+1）]=lim[2x/(1+x²)÷3]=lim[2/[3(1/x+x)]]=0
如果是
limln(1+x²)/(3x+1)=lim[ln(1/x+x)/(3+1/x)]x趋向于无穷大时为无穷大。
如果不是
limln(1+x2)/(3x+1)=limln[(1/x+2)/(3+1/x)]=ln(2/3)(x趋向于无穷大时)
如果是ln(1+x²)÷（3x+1）用罗必塔法则
lim[ln(1+x²)÷（3x+1）]=lim[2x/(1+x²)÷3]=lim[2/[3(1/x+x)]]=0
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第1个回答 2011-12-23
极限是0
lim(x→∞)ln(1+x2)/(3x+1)
=lim(x→∞)ln(x2)/(3x)
=lim(x→∞)2lnx/(3x)
=lim(x→∞)2/3*1/x
=0
lim(x→∞)ln(1+x2)/(3x+1)
=lim(x→∞)ln(x2)/(3x)
=lim(x→∞)2lnx/(3x)
=lim(x→∞)2/3*1/x
=0
第2个回答 2011-12-23
lim(x->无穷) ln(1+x^2)/(3x+1) (无穷/无穷)
=lim(x->无穷) [2x/(1+x^2)]/3
=lim(x->无穷) 2x/[ 3(1+x^2)]
=0
=lim(x->无穷) [2x/(1+x^2)]/3
=lim(x->无穷) 2x/[ 3(1+x^2)]
=0
