Discrete Maths ..2..

2013-10-17 6:27 am
5.The following is the definition for lim x→a f(x) = L:
For all real numbers ε > 0, there exists a real number δ > 0 such that for all real numbers x, if
a −δ < x < a + δ and x ≠ a then L − ε < f(x) < L + ε.
Write what it means for lim x→a f (x) ≠ L. In other words, write the negation of the definition.

6.Prove that if r and s are any two rational numbers, then (r + s)/2 is rational.

7.Prove that for all real numbers a and b, if a < b then a < (a + b)/2 < b.

回答 (2)

2013-10-17 10:43 pm
✔ 最佳答案
系統不知道在幹嘛,一直不讓我張貼,難道是怕機器人貼文嗎?我寫在記事本上再貼過來,它好像不接受,非要我另外寫幾個字。

5.
a −δ < x < a + δ <=> −δ < x - a < δ <=> |x-a| < δL − ε < f(x) < L + ε <=> − ε < f(x) - L < ε <=> |f(x) - L|< ε
lim x→a f(x) = L <=> lim x→a f(x) equal Llim x→a f (x) ≠ L <=> lim x→a f(x) "not" equal L
not (for all x in some interval => |f(x) - L| < ε) = (exist some one x in the interval => |f(x) - L| > ε)
lim x→a f (x) ≠ L<=>There exists a real numbers ε > 0,for all real number δ > 0, there exist a real numbers x , such that |x-a| < δ, then |f(x) - L| > ε
6.
r and s are any two rational numbers, so we can write r = p1/q1, s = p2/q2, where p1 p2 q1 q2 are Integer.
r + s = (p1q2 + p2q1)/(q1q2)p1q2 + p2q1 and q1q2 are integer, so r + s is rational.

7.a < b => a+a < a+b => 2a< a+b => 2a/2 < (a+b)/2 => a < (a+b)/2similarity (a+b)/2 < b
so a < (a+b)/2 < b
2013-10-17 6:53 am
7)
a < b
a + a < a + b < b + b
2a < a + b < 2b
a < (a + b)/2 < b


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