Sunday, January 17, 2010

Post # 22

So, i'm looking through my notes and limits are something i need to glue in my brain..so i'll take this time to re-type the instructions and rules so hopefully they stay there.

Limits are used to find where an x value is going on a graph. There are two different kinds of limits, limits that appraoch infinity or negative infinity or limits that approach a number.

If you are solving a limit approaching infinity, you do these things:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity

2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero

3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degrees.

Limits approaching numbers can be solved/found in a few different ways.

First, you can plug the number x is going to into the x values of the limit and see what it comes out to be. Sometimes you can do this and come out with an actual number, but other times, the bottom comes out to zero. Many people, like myself, think this means the limit is undefined, but this is not always correct. If the limit comes out to be zero, you have to use other methods to solve it. Another easy way to solve a limit is to try to break up the limit and cancle what you can, then plug in your x value. If this does not work, then you would have to plug in the limit into your calculator. If you are working with the definition of a derivative, all you have to do is take the derivative of the last term, and that is your limit!

So, hopefully this is something that people can remember..mainly myself because it always seems to slip my mind.

Now, something i'm struggling with..is it odd taht i still can't quite understand substitution. I get as far as finding u and du then i know if something's missing waht to do..but i don't understand how to plug in du into the equation..can anyone help?

3 comments:

  1. okay so for substitution you have to find the u and du .. that i can't help you with because i almost always get them backwards if it's a product rule problem type of thing..buttt
    after you find u, to find du it's just the derivative of u. after that then you integrate or solve SUBSTITUTING u and du where whatever they are in the problem. if du has something extra you add it to the beginning. then you solve and simplify. after it's fully simplified then you look at where the u is and plug in what u is where u is in the problem. and that's it . that's your answer!
    hope this helps!

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  2. Substitution can best be used for fractions and trig function (in my opinion). u is the "more complicated term" and du is the derivative of that. you can always put things in front of the intregal sign to make the problem make sence.

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  3. just plug back in u and du where they were originally in the problem. then just take the integral and solve from there. although, you don't have to plug du back in usually because it cancels out whenever you substitute something in front of it.

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