微績分lim 尋求詳細解答

lim(x->0)_(x^2[1/x])=?
Sol
lim(x->0+)_(x^2/[1/x])
=lim(y->∞)_([y]/y^2)
lim(y->∞)_(y/y^2)<=lim(x->0+)_(x^2/[1/x])<=lim(y->∞)_(y+1)/y^2
lim(x->0+)_(x^2/[1/x])=0
lim(x->0－)_(x^2/[1/x])
=lim(y->－∞)_([y]/y^2)
lim(y->－∞)_(y－1)/y^2<=lim(x->0－)_(x^2/[1/x])<=lim(y->－∞)_y/y^2
lim(x->0－)_(x^2/[1/x])=0
lim(x->0)_(x^2[1/x])=０

Since x-1 <[x] <= x for all x in R,
we have 1/x -1 < [1/x ] <= 1/x, and
x^2(1/x-1)<=x^2* 1/x.
= > x(1-x) < x^2[x] <=x......(1)
take lim x->0,
the left side of (1) =0,and
the right side of (1)=0.
Hence, lim x^2 [x]=0 by The Squeeze theorem.

2012-10-19 21:18:38 補充：
Ref. WIKI:
﹝﹞是高斯符號 參考:
http://en.wikipedia.org/wiki/Floor_and_ceiling_functions
之 Equivalences 部分第3式。
或
http://zh.wikipedia.org/wiki/高斯符號

The Squeeze theorem(夾擠原理) 參考:
http://en.wikipedia.org/wiki/Squeeze_theorem
或
http://zh.wikipedia.org/wiki/夾擠定理

lim(x→0) X^2﹝1/X﹞
=limit(x->0)[x^2/x]
=x
=0


2012-10-19 14:08:25 補充：
lim(x→0)x^2=0

limit(x->0)[1/x]=[1/0.000.....1]=1/0=∞

所以x^[1/x]=0/∞=0