If lim lim x→a g(x) = B then lim x→a 1/g(x) = 1/B
I was able to devise my own proof. However, since the book is now using its own proof for other problems, I need to understand the one given by the text:
Since lim x→a g(x) = B,
0 < | x – a | < δ1 → | g(x) – B | < |B|/2
0 < | x – a | < δ2 → | g(x) – B | < |B|^2 * 1/2 * ε
|B| = | g(x) + B - g(x) | < | g(x) | + |B|/2
|B|/2 < | g(x) | or 1/| g(x)| < 2/|B|
Let δ = min [ δ1, δ2 ]
| 1/g(x) - 1/B | = 1/|B| * 1/| g(x) | * | B - g(x) |
< 1/|B| * 2/|B| * |B|^2 * 1/2 * ε = ε
The algebra is pretty simple, but the first part is what I don't understand. Where did the values |B|/2 and |B|^2 * 1/2 * ε come from? They seem kind of arbitrary, and how do I know that those values will work for all ε? For example, since B is a constant, I can always find a ε < |B|/2.
I was able to devise my own proof. However, since the book is now using its own proof for other problems, I need to understand the one given by the text:
Since lim x→a g(x) = B,
0 < | x – a | < δ1 → | g(x) – B | < |B|/2
0 < | x – a | < δ2 → | g(x) – B | < |B|^2 * 1/2 * ε
|B| = | g(x) + B - g(x) | < | g(x) | + |B|/2
|B|/2 < | g(x) | or 1/| g(x)| < 2/|B|
Let δ = min [ δ1, δ2 ]
| 1/g(x) - 1/B | = 1/|B| * 1/| g(x) | * | B - g(x) |
< 1/|B| * 2/|B| * |B|^2 * 1/2 * ε = ε
The algebra is pretty simple, but the first part is what I don't understand. Where did the values |B|/2 and |B|^2 * 1/2 * ε come from? They seem kind of arbitrary, and how do I know that those values will work for all ε? For example, since B is a constant, I can always find a ε < |B|/2.
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There is a key component here that you don't mention, but it is absolutely critical--namely that B is not zero. Okay that said:
They are quite arbitrary, but are chosen strategically. Since lim(x->a) g(x) = B, it follows that you can find some δ1 > 0 such that
|g(x) - B| < |B|/2 provided 0 < |x - a| < δ1.
This is just the statement that for every ε > 0, there exists δ > 0 such that blah blah blah. The choice of ε here is just |B|/2.
The second statement is the the same. If ε > 0 is arbitrary, then |B|²ε/2 > 0 is some positive number. And since lim(x->a) g(x) = B, we know that for this positive number we can chose δ2 >0 sufficiently small so that
|g(x) - B| < (|B|²ε/2) whenever 0 < |x - a| < δ2.
The fact that you can find ε < |B|/2 doesn't have any real bearing on this construction. The choice of δ1 only has to work for |B|/2. It may be that δ2 would have to be significantly smaller for some specific ε, but in the end, you take δ to be the minimum so that all required inequalities hold.
The important thing is that this construction is valid without placing any restriction on ε as must be the case.
They are quite arbitrary, but are chosen strategically. Since lim(x->a) g(x) = B, it follows that you can find some δ1 > 0 such that
|g(x) - B| < |B|/2 provided 0 < |x - a| < δ1.
This is just the statement that for every ε > 0, there exists δ > 0 such that blah blah blah. The choice of ε here is just |B|/2.
The second statement is the the same. If ε > 0 is arbitrary, then |B|²ε/2 > 0 is some positive number. And since lim(x->a) g(x) = B, we know that for this positive number we can chose δ2 >0 sufficiently small so that
|g(x) - B| < (|B|²ε/2) whenever 0 < |x - a| < δ2.
The fact that you can find ε < |B|/2 doesn't have any real bearing on this construction. The choice of δ1 only has to work for |B|/2. It may be that δ2 would have to be significantly smaller for some specific ε, but in the end, you take δ to be the minimum so that all required inequalities hold.
The important thing is that this construction is valid without placing any restriction on ε as must be the case.