BOB

Bobbing for Limits. This was a short unit. When we started to talk about the graphs, I was kinda lost at first, but after a while I got it and it was okay.
Good luck on the test! o(∩_∩)o

its BOB time!

Its that time again... BOB time! Reflecting on the unit limits, at first i thought that concept of it was going to be extremely hard because its calculus and the state of mind that I'll have to learn new concepts. As the unit progress i started to understand more and more the concept became a bit more easier. Also I would like to thank my fellow classmate who helped me out to understand the unit more. Going in to the test i hope i do alright. Good luck to everyone tomorrow.

BOBby McFerrin

I just realized that I haven't blogged in a long time! Anyway, I thought that this unit, Limits, was pretty simple and quite easy to understand, that is, if you do your homework. I admit I forget to do it sometimes because we only have this class every other day. I get confused on interpreting graphs sometimes but after going over everything today, everything became clear and I understand it.

Anyway, good luck on the test tomorrow guys!

Chris' Bob

Hey It's MrSiwWy here my BOB vicariously through Grey-M since I haven't been added to the blog yet. Well, the first unit in this class didn't really present anything new to the calculus students in class or anything terribly difficult to the rest of the class. Though, I feel as if I'm bound to make tonnes of incredibly silly mistakes throughout the test tomorrow. If so, I'll probably spend a little more time with some practice problems or concept reviews during each night so everything doesn't seem so foreign anymore. Hopefully we all do well, though, it's not that difficult of a unit. I wish everyone good luck on the test tomorrow and a great night."

BOB

BOB-ing again?! WOW! Anyway, this unit took less time than I thought! It seems pretty easy so far, at least as far as I know. Who knows what I might encounter tomorrow during the test! Limits... Sounds like everything is easy because there IS a limit. Though, it gets tricky especially that part where we have to EVALUATE, and the part where we have to present evidence of continuity. Condition I. II.. III... It's interesting. We're exploring a different area in functions. I hope everyone does well tomorrow! Good luck to me... AND YOU! Study HARD. I should study now too! Goodnight everyone!

BOBBING FOR MATH

What can I say about this unit? We'll it was the first unit for Intro to Calculus and it was really boring and like bland. It was hard not to get distracted by other things. Limits isn't hard it was just writing everything out that I would forget to. Sometimes I wouldn't see the dots and thats where I lose marks. Other than the boring mechanical stuff limits was okay. That was a fast unit. We'll I guess I'm happy limits is over with.

BOB

Hey it's me GreyM! Well gotta say for being in the other calc class I'm not terribly comfortable going into this test. I'm going to chalk it up to lack of doing any hw for this class as it was pretty much last priority for me every night. We'll see if I can move it up the priority list after this test.

Bob!

Test time! Or tomorrow rather. Yep, it's that time again. This unit, on limits was a breeze. So far the unit hasn't gotten in depth yet so I don't believe that I have any problems, considering this is my second time around learning limits. I'm glad that I took this course cause despite the fact that much was a review, the second round trip opened my eyes to things that I couldn't catch during the first round. Best of luck to everyone else on the test tomorrow! Study? =)

1's - BoB

This is 1's first Blogging on Blogging post (B.o.B. or B^2).
This unit was for "limits".


I would like to comment how cool limits are. How cool is it like Mr. K said to ride on a roller coaster and all of a sudden get to a certain point and DROP straight down... This analogy was pretty cool as it related to limits. When you approach the limit of a function, the line goes into thin air! Amazingly, after this point on the graph it continued on like nothing happened at all. It is amazing in the fact that it just disappears as there is nothing in the spot...

I think what might pose problems for me is the interpreting the graphs right but other than that I will continue practice some more questions in preparation for the test.

I hope to gain some knowledge from this calculus class so that when I take it for real, I will be better prepared.

Craig's BOB

Hello everyone, this is Craig here to do my first Blogging On Blogging (B.O.B.) post
for this Calculus 45 S course.

This first unit has been a brief review for me because of my taking of the AP Calculus course, but it has been a nice refresher.

It is actually still kind of tricky because I have to learn how to use the long, tedious, drawn-out methods of solving the problems I usually do very briefly. I guess it enhances my understanding of the concepts because we actually touch on the basics compared to the fast-paced AP Calculus' version of Limits.

The pre-test was a clear example of how I need to use all of the techniques used in this class to get full marks. It is good practice and I look forward to more helpful hints from the class.

Good luck to all on the test on Wednesday =D

BOB

Well, this unit about limits went by really fast. I didn't even feel that we already finished a unit. This unit has been really easy for me ( thanks to AP Calculus). This unit has been sort of a review of the section that i learned not too long ago from calculus but is just discussed in a greater depth. The only thing that will pose problems for me in the test is evaluating using the limit theorems. I particularly don't like writing the limit notation in every step of the process, other than that i think that i should be fine. Good luck everyone on the test and study hard!!

Limits: The Scribe.. Continue or Discontinue?

Hello! On our blog, I am known as Tim-math-y, and I will be your scribe for today's lessons.

Introduction:

We started off the class with a brief discussion on our del.icio.us accounts and homework. We are to find, with effort, atleast one site that we can learn from and that can be leveled as a quality find. Then we 'tag' it with: cal45sw08, so that it will be added to our blog's bucket. Remember that it may not only aid in developing our learning outside the classroom but also, may prove to be great resources for others reading our blog.

Sweeping that discussion aside, we started off our pre-test on the unit of limits! The pre-test consisted of 5 questions in total. For those who do not know the procedures of a pre-test, it is an effective practice worth marks where a short test is written. After a set test-writing duration, we are placed into even groups where we share our answers to compile the best solutions onto one paper, as a team. This individual test paper is handed in before Mr. K reveals and explains the correct solutions.

The Pre-test:

As mentioned earlier, this Pre-test consisted of 5 questions: 2 multiple choice questions, 2 short answer questions (where work was required to show), and 1 long answer question.


The first question included an error that stumped everyone. The x^4 in the numerator was supposed to be x^2. Because of this unintentional error, this question was ommitted, as far as marks go.

However, this question could still be solved by exploring the function. This is shown on the slide. First, we notice that there is a vertical asymptote at x = 4 (Remember that when a question is asking for a limit, it is essentially looking for a horizontal asymptote).

By creating a number line, one will find that as 'x' approaches 4 from the negative side, the function goes to positive infinity. One would also find that as 'x' approaches 4 from the positive side, the function goes to negative infinity. Because of this occurrence, the limit, as 'x' approaches 4 from the positive and negative side DOES NOT EXIST.

This question is simply a give-away, as many may describe it. As 'x' approaches the value of 1 from the positive and negative side, the value is 1.

A number of groups faltered on this question simply because of the nature of previous 'short answer' questions. In the past, short answer questions were marked based on the final answer only, with a chance to earn partial marks for work shown. However, this question stated: Evaluate using the Limit Theorems, upon which many did not. This question is extremely simple yet painful. As long as you know your limit theorems, you should be fine. Listed are the limit theorems from [visual calculus]:


These are the main limit theorems we are required to know.

To start off this question, we chose to solve for the horizontal asymptote first. By dividing each term by the variable with the highest degree, we found that the horizontal asymptote y=0, when the value of 'x' approaches infinity (any number divided by infinity is extremely close to zero, therefore in this method, terms are reduced substantially).

Next we solved for the vertical asymptotes. This is done by factoring the denominator and solving for restrictions (the denominator can not equal to zero). We found the vertical asymptotes to be @ x=-9, 0.

Finally, to help visualize the graph and sketch it, a simple method of finding out the positions around the asymptotes is by creating a number line:

• As 'x' approaches -9 from the negative side, the limit is +infinity
  
• As 'x' approaches -9 from the positive side, the limit is -infinity
• As 'x' approaches 0 from the negative side, the limit is -infinity
• As 'x' approaches 0 from the positive side, the limit is +infinity

Thus, the graph can be sketched.

In the final question, we started off by running the piece-wise function through the three steps of continuity testing.

• Does f(a) exist?
• Does the limit as 'x' approaches 'a' exist?
• Does f(a) = L?

If not, the function is discontinuous.

• f(a) = f(2)
  f(2) = 2
• the limit as 'x' approaches 2 is 5
• f(2) does not equal L: 2 does not equal 5

Therefore, this piece-wise function is not continuous. By discovering this, we found that this function is a removable discontuity.

Finally, we had to sketch this piece-wise function. The graph maintains the shape of (x+3). However, it has a hole at x = 2 because there was a reduction in the factors of (x-2). Remember: when there is a reduction in factors, there is a hole at that point rather than an asymptote. Because the function of f(x) has a value of 2 @ x = 2, there is a black dot at that location.

The Conclusion:

Well that was our pre-test! To sum things up, there were multiple things that should be remembered.

• A limit as 'x' approaches a value from both sides must meet at the same point, otherwise, the limit does not exist
• Remember how to solve using the painful work of writing out all of the evaluation steps using the limit theorems
• A number line really helps in determining the shape of the function
• Remember the three steps to testing continuity
• When factors reduce a restriction in the denominator, there is a hole at that value of 'x' rather than a vertical asymptote

I hope this scribe helped any of the readers! There will be a test on wednesday so DON'T DON'T DON'T DON'T DON'T FORGET TO "BOB" !

Good luck everyone on the test! Do not forget to study either! =) Have a great night everyone.

OoOoooOOo! And the scribe for the next class will be: (Give me a sec while I find the scribe list)

.........
..........
...........

John D. !!!!!

Today's Slides: March 3

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MORE LIMITS

THURSDAY'S CLASS

We started off class with a quiz. It had two graphs and we had to find the limit of this and that. We marked them in class and went over some of them. We mainly focused on the different type of graphs that are discontinuity. The three different types of discontinuity are:


1. Removable Discontinuity:
A hole in a graph. That is, a discontinuity that can be "repaired" by filling in a single point. In other words, a removable discontinuity is a point at which a graph is not connected but can be made connected by filling in a single point.



2. Jump Discontinuity:
Jump discontinuities occur where the graph has a break in it is as this graph does. It can't be fixed or repaired so that the graph is continuous.





3. Infinite Discontinuity:
A discontinuity of a function for which the absolute value of the function can have arbitrarily large values arbitrarily close to the discontinuity. Can't be fixed either.

The formal way of telling what kind of discontinuity it is:


Then we worked on some questions in class.


We also did some work on Mr. K's favorite website for limits ( i think ). We worked together as a class to solve them and they're all on the slides that Mr. K posted up on the 28Th. It also has explanations and the homework assignments. And that was Thursday's class.
The next scribe is KIM POSSIBLE. Ha ha i totally ripped that off from your brother but he's not around.

L-I-M-I-T-S

TUESDAY'S MATH CLASS

Someone forgot to scribe for Tuesday's math class and she's making up for it by scribing for Tuesday and Thursday. Sorry. Well, on Tuesday we mainly focused on limits in a symbolic approach. The first question was....

1. Factor both the numerator and denominator. If you don't factor the numerator and denominator and go straight to substituting in the value 2 for x, you'll end up with zero in the denominator making the whole thing undefined.
   
2. Reduce
   
3. After reducing the same terms in the numerator and denominator you substitute the value two in everywhere there is an x.
   
4. Voila, you end up with the answer.

1. Rationalize the numerator because if you do then you can reduce 25 - x in the numerator and denominator.
2. After reducing, you're left with 1 over 5 + √x. You can now substitute the value 25 for x because it's in its most reduced form.
   
3. Now simplify, the √25 is 5. What is left is 1 over 10.
   




The whole thing is undefined because in the end your gonna have to substitute zero in for x and 3 over 0 is undefined. Any number over zero is undefined.




1. The numerator can be factored so that something can be reduced from the bottom.
2. After reducing to the simplest form you now can substitute the value nine for where there's an x.
   
3. The end result is negative 6.





That would have been marked wrong if it was on a test or exam. Why?
Well, in the end since it's just the notation and you've solved for it, you don't need the lim thing.


For the rest of the class, we briefly talked about horizontal and vertical asymptotes. Homework was posted in the slides.

Today's Slides: February 28

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Today's Slides: February 26

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Whoa! I can’t remember the last time I blogged for Math! But, yes, again, blogging is a very important instrument to enhance knowledge and develop self responsibility (remember the escalator video..). It is our utmost responsibility to do things ourselves even though we tend to lean on others – which means:

• do our homework
• try to blog as extensive as possible for it determines how well you know the material
• read scribe post everyday
• ask questions if necessary
• always check for misinterpretation by peers so that no one else will be confused and eventually mess up a test or exam in the future.

So, have you read the blog-online rules? Have you watched the video? Have you finished your homework? If you haven’t done so, please do.

Talking about unlimited amount of math, Mr. K introduced a very interesting approach to

UNIT 1: LIMITS

What is a limit? The first time I heard it was when I watched “Mean Girls” back then. I didn’t know what it was. Yes, 5 years after… here it comes, slowly making sense. Yes, AP CALC people, you can laugh since we’re only at the threshold of the house you have already explored three times! Here it goes.

This notation is used to express LIMITS, which means if you do not have this in every line, the entire thing is incorrect!

PROBLEM #1:

What do we first notice about this equation?

• a second degree function (x^2 : parabola) over a first degree (linear)
• a difference of square in the numerator
• if we graph this on our graphing calculator, it’s a straight line, opposing the fact that it should be some kind of a parabola. Hmmm, weird.

What happens if we factor the numerator?

• the (x-1) reduce
• we’re left with f(x) = x + 1
• graphically, it is identical to the graph we had earlier when we graphed the original equation

What if x = 1?

• Most of the students would say, “IT’S 2!” because of the equation: f(1) = 1 + 1 = 2
• However, some might disagree and say, “It’s undefined, buddy!”

The question now is WHY? Well, f(x) = x + 1 isn’t the original equation.Therefore, substituting 1 for all x gives us:

This brought the discussion about the very round number called ZERO. Usually when we divide any number by zero, we say “YOU CAN’T!!!”. It’s very hard to explain. Actually it’s pretty simple. You can’t divide by NOTHING! This follows the same curvature of the ball of wax. Anything over ZERO is undefined… it’s not two… again, it’s undefined!

We also had a discussion about 0/0 is not 1, why is it so different from 2/2 = 1?, when both 0 and 2 are numbers? Isn’t a number divided by itself equal to 1? Why is zero such an exception? Well it could mean NOTHING, INFINITY, or ZERO. It all depends on the hwo you look at it. Interesting… Mr. K, took out his “block of wood” to further discuss how you can look at something at different ways but it still refers to the same thing.

Consider SLOPE. It can be represented in three ways:

• m
• y = rise / run
• y = delta y / delta x

So, why do we have three ways to describe slope? ANSWER: because we have three ways to illustrate a function:

• equation – where ‘m’ is present in the standard form of a line (y = mx +b)
• numerically (table of values) – where we can take two ordered pairs and put them into the equation
• graphically – where we can locate two a point and use y = to find the slope.

In this case, f(1) is in an indeterminate form, which means, when x = 1, it is undefined. To prove this, we can graph the equation one more time. But this time, hit ZOOM 4, which will provide you with a much closer scrutiny at the graph.

Look at the gap on the linear equation. Isn’t it weird? Well that’s exactly what we had earlier. f(1) ix not 2. Because the point missing on the graph. Try tracing any integer greater or less than 1. It will give you the y-value but will not do it so if you enter x = 1. This squeezes out the value of 2 from both sides - >2 and <2.>

Let's take a look at a very similar problem:

As the end of class approaches its LIMIT, Mr. K, very quickly went through the laws of limits with all the mathematical operations (addition, subtraction, multiplication, and division). They are on the slides posted of Friday! It's pretty simple. It's somehow like logarithms but not really! AND AGAIN, MAKE SURE TO INCLUDE THE PROPER NOTATION FOR LIMITS OR YOUR WORK WILL LOSE A VERY FRUSTRATING AMOUNT OF MARKS!

We had a glimpse of the graphs of limits at the end of the period but since I was not sure about how it goes, I chose not to include it here. I hope Mr. K further go into details about that one in class next next class (he won't be here on Wednesday, which is the only class we have this week). This concludes my scribe post and I hope everyone had a great 3-day weekend! If you didn't, don't worry, there IS another one! YES! Anyway the next scribe will be...

K r i s t i n !

Today's Slides: February 15

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Today's Slides: February 7

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