There is no difference in the Absolute Value definition of a limit and the "within" definition that we have already learned. All the absolute value signs are used for is to show that if you're within a certain, very small, distance (which is taken the absolute value of, so that all values are positive and more easily comparable) of "X", you'll be within a similar, very small distance of "Y."

ENTRY #2: Ch. 2 “Sum It Up”

Thus far in Chapter 2, the most important thing we've learned is one sided limits and limits as x->Infinity.

I now understand that a limit is just what the value of "Y" is, or would be, at a given "X" , at a specific point on the graph, as long as the graph is a function, and has no step discontinuities.

Limits apply to derivatives and definite integrals in many ways. The most obvious of which is how you can use limits to find values on a graph which don't actually show on the graph. You can then use the limits/points to calculate derivatives or definite integrals that otherwise would be incalculatable. However, if you are told to find the derivative of a point where there is a discontinuity of some sort, you can't. This is because if there is no actual value, at that specific point, there is no rate of change, regardless of what the rest of the function is doing.

Continuity is one of the most simple things we've gone over all year. All continuity is, is the idea that you could trace a point on the graph, and for every "X" value in the given domain there is a "Y" value that exists. All the "Intermediate Value Theorem" states, is that in a given domain [A,B], there is atleast one existant "Y" value for every "X" value.

I understand almost everything we've been going over, but I'm a little foggy on the last two questions on the last test. They had to do with delta values as Y->1000, and things like that.
-Bryan

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Posted by: Period2LampLover    in: My entries

Modified on October 18, 2006 at 10:37 PM

Thus far in AP Calculus, we have learned the basics of Definite Integrals, Derivatives, and Limits.

We have learned that a derivative is the name for the "instantaneous rate of change." Which in lamest terms means, the slope of the line is at a certain point on the graph. When studying definite integrals, we learned that they are just the area below a graph between two given x values, and down to a specific Y value. Last, we learned that limits are the Y value, at a certain point of X, even if there was no actual value for Y; as if there were a removable discontinuity.

Calclulating these concepts requires different mathematical strategies.

When finding the definite integral of a function, you can use the trapazoid method to figure out the area underneath the curve of the graph. You do this by splitting the area into sections, and then by evening out the tops. To do this, you just connect the tip of each trapazoid with a line. You then use simple multiplication to figure out the area of the sections. To find out what the derivative, or instantaneous rate of change, of a function, you get points that are equal x-values away from eachother, and average the slope using the equation "Y2-Y1/X2-X1". Once you have found this number, you know the rate at which a function increases or decreases. In order to calculate the limit of a function, all you have to do is look at the graph or solve the origional equation after factoring any removable discontinuities out of it.

At this point I feel I have a good understanding of the information we have covered. However, I'm sure as we move onto the more complex ideas of Calculus I will have a lot of questions.

-Bryan

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Posted by: Period2LampLover    in: My entries

Modified on September 18, 2006 at 9:59 PM