推 monemo13 :Thus, if we choose δ≦|c|/2,we succeed in making 01/28 00:07
→ monemo13 :|x| ≧ |c|/2. 01/28 00:08
→ monemo13 :Finally, if we also require δ≦ε˙c^2/2 01/28 00:08
→ monemo13 :這邊的δ≦|c|/2 跟 δ≦ε˙c^2/2 又是怎麼成立的? 01/28 00:09
※ 引述《monemo13 (李孟李孟)》之銘言：
Prove that lim 1/x = 1/c , c≠0.
0< |x-c|< δ => | 1/x-1/c |< ε
Now
| 1/x-1/c | = | (c-x)/xc | = 1/|x|˙1/|c|˙|x-c|
The factor 1/|x| is troublesome, especially if near 0.
We can bound this factor if we keep x away from 0.
To that end, note that
|c| = |c-x+x| ≦ |c+x| + |x|
so
|x|≧ |c|-|x-c|
Thus, if we choose δ≦|c|/2,we succeed in making |x| ≧ |c|/2.
Finally, if we also require δ≦ε˙c^2/2, then
1/|x|˙1/|c|˙|x-c| < 1/|c|/2 ˙ 1/|c|˙εc^2/2 = ε
Let ε > 0 be given. Choose δ= min{|c|/2,εc^2/2}.
Then 0< |x-c|<δ implies
|1/x-1/c |= |(c-x)/xc|= 1/|x|˙1/|c|˙|x-c| < 1/|c|/2˙1/|c|˙εc^2/2 = ε
黃色的地方很不懂為什麼?...
|x| ≧ |c|/2 => 1/|x| ≦ 1/(|c|/2)
|x - c| < δ≦ε˙c^2/2 => |x-c| < ε˙c^2/2
1/|c|/2 ˙ 1/|c|˙εc^2/2 怎麼來的？
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