# Homework Help: Deleted neighborhoods, missing something about working with inequalities

1. Jul 22, 2010

### eraserhd

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.

2. Jul 23, 2010

### Gib Z

The reason your answers won't match up to the back of the book to often (and the reason why the question asks for "a" neighborhood rather than "the" ) is because these problems are about rough estimations, and just finding sufficient bounds to satisfy our requirements, and not necessarily the optimal bounds.

For example, If I asked what deleted neighborhood around x=4 will ensure 4-1 < [f(x) = x] < 4+1, whilst the optimal and most precise solution would be $0<|x-4|< 1$, $0< |x-4| < 1/2$ would suffice as well, or any other stricter inequality.

Also, your answers don't necessarily have to be pretty. For example, if we start with $$2-\epsilon < 4/x < 2+\epsilon$$, invert, multiply by 4 and subtract 2 from all, we get $$\frac{-2\epsilon}{2+\epsilon} < x-2 < \frac{2\epsilon}{2-\epsilon}$$.

Whilst that is "optimal", the result we want to imply (that $$|4/x -2 |<\epsilon$$) still holds if we make the bounds tighter and use $$0 < |x-2| < \min \left( \left|\frac{-2\epsilon}{2+\epsilon} \right|, \left|\frac{2\epsilon}{2-\epsilon} \right| \right)$$