1.lim(x->0) (ax)/(bx) = a/b2.lim(x->1) (x-1)/(x-π) = 03.lim(x->+∞) (1-1/x )^√x = lim(x->+∞) (1-1/x )^x * (1/√x)lim(x->+∞) (1-1/x )^x = e^(-1) = 1/e,1/√x -> 0原式 = 14.lim(x->∞) [1+ 1/(x+1/2) ]...謝謝。能不能再幫我看看這個(gè)lim(x-0)(x^2sin1/x)/(根號(hào)2x^2-1)lim(x-1)(絕對(duì)值x-1)/x-1已知lim(x趨于-1)（2x^2+ax+b)/(x+3)=3 求a、b設(shè)X0=1 X1+1+X0/1+X0...Xn+1=1+Xn/1+Xn證明lim(x趨于無窮）X存在，并求之。X后面的數(shù)字是下標(biāo)1. lim(x->0)(x^2sin1/x)/(根號(hào)2x^2-1) = 0 無窮小與有界函數(shù)的乘積還是無窮小2. lim(x->1)| x-1| /（x-1） 不存在， 因?yàn)樽髽O限 = -1， 而右極限 =13. lim(x->-3)（2x^2+ax+b) / (x+3) = 3 => 2 * 3^2 - 3a + b =0=> a = 6 + b/3=>2x^2+ax+b = 2(x+3)(x+b/6) lim(x->-3)（2x^2+ax+b) / (x+3)= lim(x->-3)（2(x+3)(x+b/6) / (x+3) = lim(x->-3) 2(x+b/6) = 3 =>b = 27,a = 154. X(n+1)=1 + X(n) / [1+X(n)]可以用數(shù)學(xué)歸納法證明{X(n)} 單增且 1 < X(n) < 2 由單調(diào)有界準(zhǔn)則知：{X(n)} 收斂， 設(shè)其極限為A，X(n+1)=1 + X(n) / [1+X(n)]=》A = 1 + A/(1+A) =>A = (1+√5) /2