帮我解决几道大学高数题

问题描述:

帮我解决几道大学高数题
lim(x_0)(e^ax -1)/sinbx
lim(x-1)(x-1)/(x-π)
lim(1-1/x)^根号x
lim(x-无穷){(2x+1)/(2x-1 )}^x
lim(x-无穷)(1+2^x+3^x)^1/x
会几道就帮我解几道吧,

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) ]...谢谢。能不能再帮我看看这个lim(x-0)(x^2sin1/x)/(根号2x^2-1)lim(x-1)(绝对值x-1)/x-1已知lim(x趋于-1)(2x^2+ax+b)/(x+3)=3 求a、b设X0=1 X1+1+X0/1+X0...Xn+1=1+Xn/1+Xn证明lim(x趋于无穷)X存在,并求之。X后面的数字是下标1. lim(x->0)(x^2sin1/x)/(根号2x^2-1) = 0 无穷小与有界函数的乘积还是无穷小2. lim(x->1)| x-1| /(x-1) 不存在, 因为左极限 = -1, 而右极限 =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)]可以用数学归纳法证明{X(n)} 单增且 1 A = (1+√5) /2