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eraserhd
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This is from a calculus textbook (self-study), but my problem is not with the calculus material itself. I feel I'm missing something (obvious!) from algebra. In any case:
The problem statement
Find an appropriate number L (a limit), and a deleted neighborhood N of a, such that a given e > 0,
L - e < f(x) < L + e when x is in N.
Exercise I can solve
I can solve most of the exercises, but I can't figure out some of them. Here's one I can solve:
1. f(x) = 5 - 3x, a = 2
The answer in the back of the book, which is my answer, is:
L = -1, 0 < | x - 2 | < e/3
My work is, L = 5 - 3*2 = 5 - 6 = -1
L - e < 5 - 3x < L + e
-1 - e < 5 - 3x < -1 + e
-6 - e < -3x < -6 + e
(-6 - e)/-3 > x > (-6 + e)/-3
2 + e/3 > x > 2 - e/3
e/3 > x - 2 > -e/3
0 < | x - 2 | < e/3 (to delete x=2)
Exercise I can't solve
(or at least can't get the answer in the back of the book):
3. f(x) = 4/x, a = 2
The answer in the back of the book is:
0 < | x - 2 | < 2e/(2 + e)
Here's how far I get:
L = 4/2 = 2
2 - e < 4/x < 2 + e
-e < 4/x - 2 < e
-e/4 < x - 1/2 < e/4
Now, here is where I'm missing something. I can either delete 1/2 (instead of 2),
0 < | x - 1/2 | < e/4
or, if I subtract 3/2 instead,
-e/4 - 3/2 < x - 2 < e/4 - 3/2
I can't unify the leftmost and rightmost terms because they don't have the same absolute value.
Trying really hard to get the book's answer, working from one of the above equations, I can multiply by 8/(2+e):
-e/4 < x - 1/2 < e/4
(8/(2+e)) * (-e/4) < (8/(2+e)) * (x - 1/2) < (8/(2+e))*(e/4)
-2e/(2+e) < (8x-4)/(2+e) < 2e/(2+e)
So now I'm stuck the other way because I can't reduce (8x-4)/(2+e) to (x-2).
There's more problems that seem to use the same kind of manipulation that I'm failing to find.
Thanks in advance.
The problem statement
Find an appropriate number L (a limit), and a deleted neighborhood N of a, such that a given e > 0,
L - e < f(x) < L + e when x is in N.
Exercise I can solve
I can solve most of the exercises, but I can't figure out some of them. Here's one I can solve:
1. f(x) = 5 - 3x, a = 2
The answer in the back of the book, which is my answer, is:
L = -1, 0 < | x - 2 | < e/3
My work is, L = 5 - 3*2 = 5 - 6 = -1
L - e < 5 - 3x < L + e
-1 - e < 5 - 3x < -1 + e
-6 - e < -3x < -6 + e
(-6 - e)/-3 > x > (-6 + e)/-3
2 + e/3 > x > 2 - e/3
e/3 > x - 2 > -e/3
0 < | x - 2 | < e/3 (to delete x=2)
Exercise I can't solve
(or at least can't get the answer in the back of the book):
3. f(x) = 4/x, a = 2
The answer in the back of the book is:
0 < | x - 2 | < 2e/(2 + e)
Here's how far I get:
L = 4/2 = 2
2 - e < 4/x < 2 + e
-e < 4/x - 2 < e
-e/4 < x - 1/2 < e/4
Now, here is where I'm missing something. I can either delete 1/2 (instead of 2),
0 < | x - 1/2 | < e/4
or, if I subtract 3/2 instead,
-e/4 - 3/2 < x - 2 < e/4 - 3/2
I can't unify the leftmost and rightmost terms because they don't have the same absolute value.
Trying really hard to get the book's answer, working from one of the above equations, I can multiply by 8/(2+e):
-e/4 < x - 1/2 < e/4
(8/(2+e)) * (-e/4) < (8/(2+e)) * (x - 1/2) < (8/(2+e))*(e/4)
-2e/(2+e) < (8x-4)/(2+e) < 2e/(2+e)
So now I'm stuck the other way because I can't reduce (8x-4)/(2+e) to (x-2).
There's more problems that seem to use the same kind of manipulation that I'm failing to find.
Thanks in advance.