F（x）=[(3次根号下x^2)-(2次根号下x)]除以x（x>=0）求F（x）在X=0处的极限F（x）=[(3次根号下x^2)-(2次根号下x)]除以x（x>=0）求F（x）在X=0处的极限

Answer 1

limF(X)=lim [√(x+1)-1] [√(x+1)+1][B]/{[^3√(x+1) -1][√(x+1)+1][B]}
=lim[x+1-1][B]/[1+x-1][[√(x+1)+1]=limB/A=3/2