semidevil
- 156
- 2
I'm not understanding the definition. For any epislon > 0, there exists a delta > 0, st if x belongs to A and 0 < |x - c| < delta, then |f(x) - L| < epsilon.
ok, so to solve a problem, it means that I assume that epsilon is greater then 0, and then, depending on the problem I need to pick a delta that is greater then 0, and it has to satisfy the 0 < |x - c| < delta...and then, it will satisfy |f(x) - L | < epsilon right?
I"m not quite understanding it. So here is the example from book
prove that the lim b from x to c = b.
So first, the book picks f(x) = b. So for any epislon > 0, they let delta be 1. this means if 0 < | x - c | < 1, then |f(x) - b| = |b - b| = 0, which is < epislon.
I don't follow this quite. how/why did they choose f(x) = b? I know it makes sense because b - b = 0, and then it is less then epsilon, so QED, but doesn't it mean you can do the same thing to all limits?
ok, so to solve a problem, it means that I assume that epsilon is greater then 0, and then, depending on the problem I need to pick a delta that is greater then 0, and it has to satisfy the 0 < |x - c| < delta...and then, it will satisfy |f(x) - L | < epsilon right?
I"m not quite understanding it. So here is the example from book
prove that the lim b from x to c = b.
So first, the book picks f(x) = b. So for any epislon > 0, they let delta be 1. this means if 0 < | x - c | < 1, then |f(x) - b| = |b - b| = 0, which is < epislon.
I don't follow this quite. how/why did they choose f(x) = b? I know it makes sense because b - b = 0, and then it is less then epsilon, so QED, but doesn't it mean you can do the same thing to all limits?