微積分 w1 ex02-01q3
(單選題)
Let f (x) = 2x, c = 1, = 0.1, L = 2, pick up an appropriate positive number 8 from the following options
such that if 0 < |x − 1| < 8, then |f(x) − L| = |2x − 2| < € = 0.1.
○ 8 = 4
O
8 = 0.049
8= 0.051
8 = 0.1
解答:
"8=0.049
"
詳解:
Observe that | 2x-21 = 2|x1| < 0.1 =
>>
|x1|< 0.1 =
0.05.
2
Hence for &= 0.1, we can take any positive number 8≤ 0.05, such that if 0 < |x − 1| < 8, then
|f(x) – L| = |2x-2|
2x 128≤ 2 0.05 = 0.1 = ε.
