本文共 1919 字，大约阅读时间需要 6 分钟。
解析：根据 x − 1 < [ x ] ⩽ x x-1<[x] \leqslant x x−1<[x]⩽x，有
10 x − 1 < [ 10 x ] ⩽ 10 x \frac{10}{x}-1<\left[\frac{10}{x}\right] \leqslant \frac{10}{x} x10−1<[x10]⩽x10 于是 { x > 0 ⇒ 10 − x < x ⋅ [ 10 x ] ⩽ 10 x < 0 ⇒ 10 − x > x ⋅ [ 10 x ] ⩾ 10 \left\{\begin{array}{l} {x>0 \Rightarrow 10-x<x \cdot\left[\frac{10}{x}\right] \leqslant 10} \\ {x<0 \Rightarrow 10-x>x \cdot\left[\frac{10}{x}\right] \geqslant 10} \end{array}\right. { x>0⇒10−x<x⋅[x10]⩽10x<0⇒10−x>x⋅[x10]⩾10 故可得 I = lim x → 0 [ 10 x ] = 10 I=\lim \limits_{x \rightarrow 0}\left[\frac{10}{x}\right]=10 I=x→0lim[x10]=10解析：当 x → ( 1 n ) − x \rightarrow\left(\frac{1}{n}\right)^{-} x→(n1)−，有：
1 n + 1 < x < 1 n , n < 1 x < n + 1 \frac{1}{n+1}<x<\frac{1}{n}, \quad n<\frac{1}{x}<n+1 n+11<x<n1,n<x1<n+1 故 [ 1 n ] = n [\frac{1}{n}]=n [n1]=n,所以 lim x → ( 1 n ) − f ( x ) = lim x → ( 1 n ) x ⋅ [ 1 x ] = 1 \lim _{x \rightarrow\left(\frac{1}{n}\right)^{-}} f(x)=\lim _{x \rightarrow\left(\frac{1}{n}\right)} x \cdot\left[\frac{1}{x}\right]=1 x→(n1)−limf(x)=x→(n1)limx⋅[x1]=1 当 x → ( 1 n ) + x \rightarrow\left(\frac{1}{n}\right)^{+} x→(n1)+，有 1 n < x < 1 n − 1 , n − 1 < 1 x < n \frac{1}{n}<x<\frac{1}{n-1}, n-1<\frac{1}{x}<n n1<x<n−11,n−1<x1<n，故 [ 1 n ] = n − 1 [\frac{1}{n}]=n-1 [n1]=n−1，所以 lim x → ( 1 n ) + f ( x ) = lim x → ( 1 n ) + x ⋅ [ 1 x ] = n − 1 n < 1 \lim _{x \rightarrow\left(\frac{1}{n}\right)^{+}} f(x)=\lim _{x \rightarrow\left(\frac{1}{n}\right)^{+}} x \cdot\left[\frac{1}{x}\right]=\frac{n-1}{n}<1 x→(n1)+limf(x)=x→(n1)+limx⋅[x1]=nn−1<1 故 x = 1 n ( n = 2 , 3 , ⋯ ) x=\frac{1}{n}(n=2,3, \cdots) x=n1(n=2,3,⋯)是 f ( x ) f(x) f(x)的跳跃间断点，所以B正确。转载地址：http://ivndz.baihongyu.com/