Quote from Yoorah;559628:
Ah, fun stuff. It's been years since I've done this. xD
But what's the deal with the "up and down" function = no limit exists? O_o I don't recall hearing about that. Teach me, great math master! o/
There are four types of discontinuities, right? Asymptotic, Point, Jump, and Oscillation. Most people don't hear about the fourth one but it exists in odd functions (traditional example being sine(1/x) pics below). I'm no math major (not to mention good at math either...) but I believe it has to do with the fact that the function becomes more and more wobbly as you get closer and closer to the limit. In other words, you can't discern an appreciable limit no matter how close you are to the limit.
[Image: http://www.math.washington.edu/~conroy/general/sin1overx/g1Ann.png]
[Image: http://www.math.washington.edu/~conroy/general/sin1overx/g4.png]
You can prove that limits exist within weird functions like this using something called the Squeeze Theorem. In layman's terms, think of this as a curvy rode where you can approach from both the left and the right directions. You don't know what kind pathway the rode is between the two points where you started but you do know that the road
is connected at
some point in
some way. In other words, the road just don't have a chunk missing from it in the middle and the route is continuous despite it being incredibly curvy. From this, it can be postulated that the roads
will meet at a point because the left and the right have to be joined somewhere for it to be continuous. This point of union is thus our limit.
Here's the more formal statement as quoted from Wikipedia.
"Statement
The squeeze theorem is formally stated as follows.
Let I be an interval having the point a as a limit point. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have:
g(x) ≤ f(x) ≤ h(x)
And also suppose that:
g(x) and h(x) both approach L as x approaches a,
Then f(x) must also approach L as x approaches a.
Then
The functions g(x) and h(x) are said to be lower and upper bounds (respectively) of f(x).
Here a is not required to lie in the interior of I. Indeed, if a is an endpoint of I, then the above limits are left- or right-hand limits.
A similar statement holds for infinite intervals: for example, if I = (0, ∞), then the conclusion holds, taking the limits as x → ∞.
Proof. From the above hypotheses we have, taking the limit inferior and superior:
[Image: http://upload.wikimedia.org/math/7/8/c/78c9a4a7b4e6d3d21f295f627604eaf7.png]
so all the inequalities are indeed equalities and the thesis immediately follows."
Hope this makes sense... I truly hated this course in College because the tests almost exclusively involved proving **** like this.