Anyway, these are the main things we did in class today:
1. Warm up
2. Quiz review
3. "NAG" this limit
I have links at the bottom for some things I needed to brush up on, so if you need to review some of the things we did today that we learned last year, see if there's anything that'll help.
Warm Up
The name of the warm up today was "Intermediate Forms", and consisted of two problems with four parts each. The first problem was to find f(x) and g(x) such that f(c) and g(c) equalled 0, and:
a) = 0
b) = .gif)
c) = 17 (or some other random number)
d) does not exist.
Kyle and Marcel found part a with a piecewise function, defining f(x) as .gif)
and g (x) as {.gif)
&.gif)
}. When graphed, we found that this satisfied all of the requirements, but Mr. O'Brien had a different way of doing it.
.gif)
and g (x) as {.gif)
&.gif)
}. When graphed, we found that this satisfied all of the requirements, but Mr. O'Brien had a different way of doing it. Mr. O'Brien suggested making it a little bit simpler and starting with an equation we knew would be easy to work with: f(x) was set equal to .gif)
with x approaching 1. This would make f(c) equal zero, so that's all taken care of. Next, we tried to find a way that we could essentially keep whole shabang equal to while still having g(c) equal zero. So, we started off with this:
.gif)
with x approaching 1. This would make f(c) equal zero, so that's all taken care of. Next, we tried to find a way that we could essentially keep whole shabang equal to while still having g(c) equal zero. So, we started off with this: .gif)
From there, we sort of snowballed ideas ranging from absolute values to making g(x) equal to .gif)
to some crazy ideas that seemed waaaay too complicated to work. So after a few more minutes of mulling, Mr. O'Brien put up a really simple equation that made all the super complicated equations we had look silly:
.gif)
to some crazy ideas that seemed waaaay too complicated to work. So after a few more minutes of mulling, Mr. O'Brien put up a really simple equation that made all the super complicated equations we had look silly:.gif)
This way, after you cancel, you're left with ; even without graphing, you can tell that as .gif)
.
; even without graphing, you can tell that as .gif)
.This sort of helped us figure out b, c, and d. We stuck with having 1 as the value for c, and after some discussion we ended up with the following solutions. I have videos for the actual work through of the problems that I'll link to at the bottom of the page, but they all go with the same concept as part a: find a way to satisfy the limit, then play with the equation to keep it the same but have it satisfying all of the requirements.
b) .gif)
, .gif)
.gif)
, .gif)
.gif)
, .gif)
d) .gif)
, .gif)
.gif)
, .gif)
It's pretty easy to see why all these would work, but if you were a little fuzzy on them yesterday, it may be a good idea to look at the videos at the bottom :]
Part two of the warm up was a lot easier since .gif)
for all of them. The best way to think about these, then, was to imagine a really really really big number for x in the equation. That way, you could see what parts of it were really important. Part a, for example, was looking for the limit of .gif)
. If you think of just a really huge number for x, it eventually becomes the natural log of 1 over a huge huge huge number, and a little number divided by a big number is a little number itself. So, all the stuff in the parentheses essentially equals zero. The natural log of zero becomes the limit, which comes out to negative infinity.
.gif)
for all of them. The best way to think about these, then, was to imagine a really really really big number for x in the equation. That way, you could see what parts of it were really important. Part a, for example, was looking for the limit of .gif)
. If you think of just a really huge number for x, it eventually becomes the natural log of 1 over a huge huge huge number, and a little number divided by a big number is a little number itself. So, all the stuff in the parentheses essentially equals zero. The natural log of zero becomes the limit, which comes out to negative infinity. Part b asked about limit of .gif)
. This is also a good one to just think about, because if x is getting really big, it'll be a little number (since natural logs don't ever get that big) divided by a big number, it'll just be zero. So, the limit is zero.
.gif)
. This is also a good one to just think about, because if x is getting really big, it'll be a little number (since natural logs don't ever get that big) divided by a big number, it'll just be zero. So, the limit is zero.Part c followed suit and had us find the limit of .gif)
. e basically equals 3, at least for this problem, so 3 to a really really really big number is going to be bigger than that really really big number squared. Mr. O'Brien had us think about last year's problem about folding a piece of paper in half fifty times. If you could do it, you'd have a stack big enough to almost reach the sun! If you apply that same logic to this problem, you can see why three raised to the millionth power is way bigger than a million squared. The limit, then, became a huge number over a little number, or infinity.
.gif)
. e basically equals 3, at least for this problem, so 3 to a really really really big number is going to be bigger than that really really big number squared. Mr. O'Brien had us think about last year's problem about folding a piece of paper in half fifty times. If you could do it, you'd have a stack big enough to almost reach the sun! If you apply that same logic to this problem, you can see why three raised to the millionth power is way bigger than a million squared. The limit, then, became a huge number over a little number, or infinity.Quiz Review
After the warm up (which actually took like forty-five minutes, so it wasn't really a warm up anymore), we went on to look at the quiz we took last week. Generally, people did really well. There were a couple things we needed to be careful of, though:
- When you work through a problem algebraically, be sure you don't accidentally write that the limit of a problem is equal to a simplified version of the function. Limits are equal to limits, functions are equal to functions.
- If you don't recognize a fraction, try to see if you can find what its real value is! This was especially applicable to number five, where the answer was
.gif)
but some people left it as ~ 1.72 or something. I was one of those people, not gonna lie, so for me at least it's something to keep in mind.
.gif)
So that's about it for the quiz! If you have any specific questions about the quiz, it'd be a good idea to email Mr. O'Brien so you can really get a handle on the problem.
"NAG" This Limit
This was the part where it got really tricky.. I thought so at least. We started off with a function whose limit we were told is the base of all calculus trigonometry: .gif)
as .gif)
. Ummm okay, talk about intimidating. But it wasn't really that bad. "NAG" stands for "Numerically, Algebraically, & Graphically" which were the three ways we sort of interpreted this problem. I'll just tell you right now that the limit is 1.
.gif)
as .gif)
. Ummm okay, talk about intimidating. But it wasn't really that bad. "NAG" stands for "Numerically, Algebraically, & Graphically" which were the three ways we sort of interpreted this problem. I'll just tell you right now that the limit is 1. So numerically, you have to sort of substitute in numbers. Think of the sin of a unit circle: the x values all rotate between positive and negative one. So, as x gets closer and closer and closer to zero, sin(x) gets closer and closer and closer to negative or positive one. If x is negative, the sin will also be negative, and two negatives make a positive. So, the limit is always positive one.
The second way we looked at it was probably the most helpful for me. If you look at the graph of the problem, it all makes more sense.
I wish that was a little bigger, but it's just the graph of sin(x)/x. As you can see, as you get closer to zero from each side, it approaches positive one. It doesn't, however, actually have a value at x = 0 since it breaks the function... You can't see that in the graph, but it's definitely true. I'm pretty sure it's the easiest way to look at the function, so this is the one I'd stick with.
The third way we looked at it was algebraically, which opened up a whole new can of worms towards the end of class. Mr. O'Brien introduced this thing called "The Squeeze (or Sandwich) Theorem". It basically says that if f(x) is less than h(x) which is less than g(x), and if f(x) has the same limit as g(x), then then limit of h(x) is the same as f(x) and g(x). at a certain interval.
I'm not super good at explaining things, but Mr. O'Brien showed us one really good way of looking at it.
Think of this circle & triangles:
The angle BAC is going to be called theta. So, you can see that the green triangle is less than the orange one, and the orange one is less than the blue one. Or, h(x) <>
h(x): cos(theta)
f(x): sin(x)/x
g(x): 1
According to the Squeeze Theorem, since h(x) and g(x) both have a limit of 1, f(x) must have one as well. I'm still not super clear on how it all works, so I found some pages that have a lot of good information:
If you wanted to practice some problems, there's also this page that'll probably explain this a lot better than I did.. hahaha sorry!
Reference Pages
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