limx趋向于无穷(1/(√x∧2+1)+1/(√x∧2+2）+.+1/(√x∧2+x))用夹逼准则证明等于1

limx趋向于无穷(1/(√x∧2+1)+1/(√x∧2+2）+.+1/(√x∧2+x))用夹逼准则证明等于1 计算极限：limx趋向于无穷,[（5x^2+1）/(3x-1)]sin1/x 求极限 limx趋向无穷(2x+3/2x+1)∧x+1 limx趋向于无穷x²-1/2x²-x-1求极限 limx趋向于无穷时4x^3-2x+8/3x^2+1 求极限,limx趋向于正无穷ln(1+3^x)/(1+2^x) 求limx^2（sin1/x)/根号（2x^2-1）在x趋向于正无穷的极限 求极限：lim[√（2X+1）-3]/[√(x-2)-√2] x趋向于4?limx/[1-√(1+x)] x趋向于0 （ 极限limx趋向于负无穷,根号下(4X^2+x)*ln(2+1/x)-2ln2*x limx趋向于正无穷,根号x+1减根号x除以根号x+2减根号x limx趋向于无穷大(x^2+1/2x+1)sin4/x等于多少 求极限limx趋向无穷 ((1-2x) /(2-2x))^x