Question

几道简单高数求极限问题求大神解答！！！ 要详细过程！！！

Answer 1

1. lim[(-2)^n + 3^n]/[(-2)^(n+1) + 3^(n+1)]

   =lim[(-2/3)^n + 1]/[(-2/3)^n *(-2) + 3]    注：分子、分母同除以 3^n

   =lim(0+1)/[0*(-2) + 3]

   =1/3

2. lim√n *[√(n+1) - √(n-1)]        注：分子、分母同乘以 [√(n+1) + √(n-1)]

   =lim√n * [√(n+1) -√(n-1)]*[√(n+1) +√(n-1)]/[√(n+1) + √(n-1)]

   =lim √n * [(n+1) - (n-1)]/[√(n+1) + √(n-1)]

   =lim √n * 2/[√(n+1) + √(n-1)]

   =lim 2/[√(1+1/n) + √(1-1/n)]        注：分子、分母同除以 √n

   =lim 2/[√(1+0) + √(1-0)]

   =1

3. 因为 1 - 1/n^2 = (n^2-1)/n^2 = (n+1)/n * (n-1)/n，所以，

   lim(1-1/2^2)(1-1/3^2)……(1-1/n^2)

   =lim(1/2*3/2)(2/3*4/3)……[(n-1)/n*(n+1)/n]

   =lim[1/2*(n+1)/n]

   =lim1/2*(1+1/n)

   =lim1/2*(1+0)

   =1/2

4. lim[1/(1+2) + 1/(1+2+3) + ……+1/(1+2+3+……+n)]

   =lim{2/(2*3) + 2/(3*4) + ……+ 2/[n(n+1)]}

   =2*lim{(1/2 - 1/3) + (1/3 - 1/4) + ……+ [1/n - 1/(n+1)]}

   =2*lim[1/2 - 1/(n+1)]

   =2*lim(1/2 - 0)

   =1

追问

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