Question

求极限，最好有详细过程，谢谢

Answer 1

1. lim x*[ln(x+1)-lnx]

   =limx*ln[(x+1)/x]

   =limx*ln(1+1/x)

   =lim ln(1+1/x)^x

   =ln lim(1+1/x)^x

   =lne = 1        注：当 x→∞时，lim(1+1/x)^x = e。这是一个基本极限公式

2. lim[√(x^2+x) -√(x^2-x)]

   =lim[√(x^2+x) -√(x^2-x)]*[√(x^2+x) +√(x^2-x)]/[√(x^2+x) +√(x^2-x)]

   =lim[(x^2+x) -(x^2-x)]/[√(x^2+x) + √(x^2-x)]

   =lim 2x/[√(x^2+x) +√(x^2-x)]

   =lim2/[√(1+1/x) +√(1-1/x)]

   =lim2/[√(1+0) +√(1-0)]

   =1

3.  lim[√(x+1) -√(3-x)]/(x-1)

   =lim[√(x+1) -√(3-x)]*[√(x+1) + √(3-x)]/{(x-1)*[√(x+1) +√(3-x)]}

   =lim[(x+1) -(3-x)]/{(x-1)*[√(x-1) +√(3-x)]}

   =lim2(x-1)/{(x-1)*[√(x+1) + √(3-x)]}

   =lim 2/[√(x+1) +√(3-x)]

   =lim 2/[√(1+1) + √(3-1)]

   =lim 2/(2√2)=√2/2

4. 设 t = 3/x，则当 x →∞时，t→0。那么，

   lim(1+3/x)^(2x+1)

   =lim(1+t)^(6/t +1)

   =lim(1+t)^(6/t) * (1+t)

   =lim[(1+t)^(1/t)]^6 *lim(1+t)

   =e^6 * 1 = e^6

追问

为什么ln（x＋1）-lnx，能变成ln（x＋1）／x。公式？

追答

这是对数函数的基本性质：
ln(x*y) = lnx + lny
ln(x/y) = lnx + ln(1/y) = lnx + lny^(-1) = lnx + (-1)*lny = lnx - lny