Explain the following statements:
a. A jump NEVER has a limit
b. An asymptote (infinite) might have a limit
c. A removable ALWAYS has a limit
Monday, August 30, 2010
Sunday, August 29, 2010
blog numero uno
its alaina :)
so in the last two weeks, we reviewed limits, continuity and discontinuity.
-if something is continuous, that implies that the graph can be drawn without picking up your pen
there are four types of discontinuity- you have to pick up your pen to draw
types of discontinuity
1. vertical asymptotes
2. removable- "hole"
3. jump
4. oscillation--don't deal with these
most problems are set up as telling you to find something,
for example, find the "colored in y-value" or find all discontinuities and name them,
and then give you a graph of f(x) formula which you must then graph or plug a number into to find the discontinuity, if any.
if a graph has a jump or a removable, discontinuities, the limit does not exist- DNE.
as i was doing this blog, i started thinking about "discussing" continuity and what exactly it means, and i've come to the conclusion that i don't exactly know. So in class tomorrow, i will probably as a question about it.
so in the last two weeks, we reviewed limits, continuity and discontinuity.
-if something is continuous, that implies that the graph can be drawn without picking up your pen
there are four types of discontinuity- you have to pick up your pen to draw
types of discontinuity
1. vertical asymptotes
2. removable- "hole"
3. jump
4. oscillation--don't deal with these
most problems are set up as telling you to find something,
for example, find the "colored in y-value" or find all discontinuities and name them,
and then give you a graph of f(x) formula which you must then graph or plug a number into to find the discontinuity, if any.
if a graph has a jump or a removable, discontinuities, the limit does not exist- DNE.
as i was doing this blog, i started thinking about "discussing" continuity and what exactly it means, and i've come to the conclusion that i don't exactly know. So in class tomorrow, i will probably as a question about it.
Justin's Blog #1
This is Justin, unfortunately I have to do my blog on Dustin's account because for some reason I didn't get the invitation on any of my emails, but anyways...
for the first 10 days of schools we talked about limits.
In definition:
"In mathematics, the concept of a "limit" is used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concept of a limit allows one to, in a complete space, define a new point from a Cauchy sequence of previously defined points. Limits are essential to Calculus( and mathematical analysis in general) and are used to define continuity, derivatives, and integrals. The concept of the limit of a function is further generalized to the concept of topological net, while the limit of a sequence is closely related to limit and direct limit in theory. In formulas, limit is usually abbreviated as lim as in lim(an)=a or represented by the right arrow(->) as in an -> a" (wikipedia)
To find the limit, plug in for x to find a y value(which is a limit). Most of the time, directly plugging in gets a 0 somewhere, so you have to go further...by factoring, squeeze, graphing, chart, etc.
The chart simply works by recording the number the limit is directly approaching:
Ex. lim , you create a chart with the values that approach zero
x->0
-.1|-.01|-.001|0|.001|.01|.1
for the first 10 days of schools we talked about limits.
In definition:
"In mathematics, the concept of a "limit" is used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concept of a limit allows one to, in a complete space, define a new point from a Cauchy sequence of previously defined points. Limits are essential to Calculus( and mathematical analysis in general) and are used to define continuity, derivatives, and integrals. The concept of the limit of a function is further generalized to the concept of topological net, while the limit of a sequence is closely related to limit and direct limit in theory. In formulas, limit is usually abbreviated as lim as in lim(an)=a or represented by the right arrow(->) as in an -> a" (wikipedia)
To find the limit, plug in for x to find a y value(which is a limit). Most of the time, directly plugging in gets a 0 somewhere, so you have to go further...by factoring, squeeze, graphing, chart, etc.
The chart simply works by recording the number the limit is directly approaching:
Ex. lim , you create a chart with the values that approach zero
x->0
-.1|-.01|-.001|0|.001|.01|.1
Blog #1
Yea...this is Dustin. The past 2 weeks, we learned a lot of nonsense about limits. So, i'll explain how limits work and stuff.
Ex. Lim x+2
x->2
For this simple problem, you would just plug in 2 for all appearances of "x".
You would then receive the simple problem 2+2, which is like 4 or something.
Limits also can be more complicating than just plugging in numbas. You might have to factor like this one here:
Ex. Lim (x^2+x-2)/(x+2)
x->-2
For this problem if you were to simply plug in the -2 you would then have to divide by zero, which is kinda tough to do. So instead, we will factor the top and get:
((x+2)(x-1))/(x+2)
Which simplifies to:
(x-1)
Then we plug in the -2 and get:
-2-1
Which is (-3).
Yea, so that's basically limits. You just plug in to the equation. If you get a whole number, then that's your answer, but if you get something divided by zero, you have to either factor, squeeze, create a chart, or some other over complicated method.
And um, something I don't understand is how to find the difference of 2 cubes, so if someone is rather smart could you, um, maybe help me or something like that. I guess that's it, and I hope that's enough words.
Ex. Lim x+2
x->2
For this simple problem, you would just plug in 2 for all appearances of "x".
You would then receive the simple problem 2+2, which is like 4 or something.
Limits also can be more complicating than just plugging in numbas. You might have to factor like this one here:
Ex. Lim (x^2+x-2)/(x+2)
x->-2
For this problem if you were to simply plug in the -2 you would then have to divide by zero, which is kinda tough to do. So instead, we will factor the top and get:
((x+2)(x-1))/(x+2)
Which simplifies to:
(x-1)
Then we plug in the -2 and get:
-2-1
Which is (-3).
Yea, so that's basically limits. You just plug in to the equation. If you get a whole number, then that's your answer, but if you get something divided by zero, you have to either factor, squeeze, create a chart, or some other over complicated method.
And um, something I don't understand is how to find the difference of 2 cubes, so if someone is rather smart could you, um, maybe help me or something like that. I guess that's it, and I hope that's enough words.
Blog #1--Stephen Ledbetter
Okay, so this is my first blog for this year in AP Calc. I must say that the whole site looks kinda different than last year...it looks good. So, we started AP Calculus this year, and for the past two weeks we've been doing nothing but limits. Limits limits limits! But we're almost done with 'em. Limits are easy if you know how to do all the "stuff" necessary to "solve" them. And these last two weeks have been just a basic review of the last part of Advanced Math last year (Pre-Calculus). Anyway, let me get to the actual math part of the blog :/
I'll explain in my own words how to solve a limit with a zero in the denominator by way of making a chart.
If you happen to come across a limit problem where the denominator equals 0, you can make a chart to find the limit as x approaches a said number from the left and right sides of the graph (number that x approaches).
You use the values of 0.001, 0.01, and 0.1 to determine these values.
On the left side, (from left to right) the values would be (number - .1, # - .01, # .001). On the right side (from left to right) the values would be (number + .001, # + .01, # + .1).
You plug the equation of the limit into y= on your calculator, then plug the decimal values into the x values on the table. (If you plug in the actual number that x approaches, you should see "ERROR" on the table.)
Then, you just see what the limit is approaching from the left (left to right) to the median number, and then what the limit is approaching from the right (right to left) to the median number. If they match, that is the limit. If they don't, then it is said it does not exist (DNE).
Example:
the limit as x approaches 4 of ((1)/(x-4))
You plug the equation into the calculator.
Now plug in
3.9 = -10
3.99 = -100
3.999 = -1000
4 = ERROR
4.001 = 1000
4.01 = 100
4.1 = 10
You can see that as x approaches 4 from the left, the limit is (negative)infinity.
As x approaches 4 from the right, the limit is infinity.
They do not match, therefore the limit does not exist (DNE).
I really don't understand how to discuss continuity. I'm still a little iffy on removables and jumps and I can't seem to grasp the concept of them. Maybe a definition and a couple of examples of them could help. Thanks.
Blog 1. This....is.....SEWIOUS!!
Haha, this blog stuff ain't gonna be pretty, but WHAT THE FIRETRUCK, might as well do it anyway.
This is Connor, I think, possibly, hopefully.....guess you'll never know!! muahahahaha
XD
I really hope the blogs are supposed to be at least 150 words long, at least, because I've totally forgotten everything that we got told during the first 2 weeks of the school year.
XD
How about them Saints the other night?! They pretty much OWNED the San Diego Chargers in the Dome the other day. And that guy with the 76 yard TD should definitely be in competition to take the RB position soon.
Well, guess it's time to get to something I loathe with a passion......math......
I think I'mma talk about limits, since thats most of what we've done in the past weeks.
Better yet, I'll talk about the discontinuities of graphs and how they mostly don't have limits.
A discontinuity occurs when a graph doesn't have a definite limit.
You can figure out when you have a discontinuity if you can't plug whatever x approaches into the equation of the graph.
example: lim x^2/x-9
x->9
you can't plug 9 into the equation, so you most likely have a discontinuity
The question is, what kind of discontinuity is it?
Well my friends, it's time I introduce you to our 3 types of discontinuities:
1) Removable: This happy-go-lucky fellow is one who is housed on the actual graph, but he is not filled in on the talk of the graph, but he is considered a limit at times, usually when he goes into Ninja mode and sneaks into the graph.
2) Jump: Ahhh, the magic of science helps to create this type of discontinuity. It approaches a point from 2 different directions at different places! That's science at it's best. It uses the "warp" method of transportation to hit a certain point on the graph, and then it pops up somewhere else on the same x point, but at a different y point. It's a freaking genius man. And you can't see it move from one point to the other, so the two points don't connect at all. Sometimes he works with the ninja-like removables, but that's on rare occasions.
3) Vertical Asymptotes: Dude, these motherfudgers are so awesome, they don't even touch a point, they just approach them, and that point is never heard from again... Usually, they consider themselves too high classed, or too low classed, to be seen with the other types of discontinuities. They end up calling themselves either the Infinites or the Negative Infinites. Those two biker gangs just dodge everything else on the plain and stay to themselves. They aren't that violent. You should meet them some time. They like classic muscle cars.
And that was the story all about how their limits got flip turned upside down. And I would like to take a minute, just sit right there, I'll tell you how I became the prince of a town called Bel-Air.
:P
This is Connor, I think, possibly, hopefully.....guess you'll never know!! muahahahaha
XD
I really hope the blogs are supposed to be at least 150 words long, at least, because I've totally forgotten everything that we got told during the first 2 weeks of the school year.
XD
How about them Saints the other night?! They pretty much OWNED the San Diego Chargers in the Dome the other day. And that guy with the 76 yard TD should definitely be in competition to take the RB position soon.
Well, guess it's time to get to something I loathe with a passion......math......
I think I'mma talk about limits, since thats most of what we've done in the past weeks.
Better yet, I'll talk about the discontinuities of graphs and how they mostly don't have limits.
A discontinuity occurs when a graph doesn't have a definite limit.
You can figure out when you have a discontinuity if you can't plug whatever x approaches into the equation of the graph.
example: lim x^2/x-9
x->9
you can't plug 9 into the equation, so you most likely have a discontinuity
The question is, what kind of discontinuity is it?
Well my friends, it's time I introduce you to our 3 types of discontinuities:
1) Removable: This happy-go-lucky fellow is one who is housed on the actual graph, but he is not filled in on the talk of the graph, but he is considered a limit at times, usually when he goes into Ninja mode and sneaks into the graph.
2) Jump: Ahhh, the magic of science helps to create this type of discontinuity. It approaches a point from 2 different directions at different places! That's science at it's best. It uses the "warp" method of transportation to hit a certain point on the graph, and then it pops up somewhere else on the same x point, but at a different y point. It's a freaking genius man. And you can't see it move from one point to the other, so the two points don't connect at all. Sometimes he works with the ninja-like removables, but that's on rare occasions.
3) Vertical Asymptotes: Dude, these motherfudgers are so awesome, they don't even touch a point, they just approach them, and that point is never heard from again... Usually, they consider themselves too high classed, or too low classed, to be seen with the other types of discontinuities. They end up calling themselves either the Infinites or the Negative Infinites. Those two biker gangs just dodge everything else on the plain and stay to themselves. They aren't that violent. You should meet them some time. They like classic muscle cars.
And that was the story all about how their limits got flip turned upside down. And I would like to take a minute, just sit right there, I'll tell you how I became the prince of a town called Bel-Air.
:P
Blog 1.
I don't know if we have to say this or not but this is Lindsey Rodrigue.
So the first week of Calculus was not as hard as I was expecting it to be, haha. I pretty much understood everything we learned but on every section there were one or two problems that I had trouble with. The first thing we learned was section 1.2, and it was just about using a table to find limits. When using a table to find limits, they either give you x value numbers already and you plug them into your calculator, or you have to take the x approaching number and subtract or add it by these numbers: -.1,-.01,-.001,.001,.01,.1. Type the problem into y= (on your calc.) Then press 2nd Graph and a table should appear. Then you plug into the table, and you get your answers. Whatever the number approaches, that is the limit. **When a problem asks you to do something numerically, it means create a table.
Ex. 1:
Lim
X->4 x-4/x^2-3x-4
x: 3.9 3.99 3.999 4.001 4.01 4.1
f(x): .20408 .2004 .20004 .19996 .1996 .19608
The limit would be x=.2
**Alright now for something I didn’t understand. Something that I thought was kind of confusing was infinite limits. The ones that I’m having trouble with are probably really simple, but I don’t get exercise 1-4 on 1.5. I don’t understand how you find if x approaches infinity or –infinity. Also, I don’t know what you’re supposed to plug in for problems 37 and 39. (When it’s like numbers from the right or left) If anyone could help me, I would appreciate it.
So the first week of Calculus was not as hard as I was expecting it to be, haha. I pretty much understood everything we learned but on every section there were one or two problems that I had trouble with. The first thing we learned was section 1.2, and it was just about using a table to find limits. When using a table to find limits, they either give you x value numbers already and you plug them into your calculator, or you have to take the x approaching number and subtract or add it by these numbers: -.1,-.01,-.001,.001,.01,.1. Type the problem into y= (on your calc.) Then press 2nd Graph and a table should appear. Then you plug into the table, and you get your answers. Whatever the number approaches, that is the limit. **When a problem asks you to do something numerically, it means create a table.
Ex. 1:
Lim
X->4 x-4/x^2-3x-4
x: 3.9 3.99 3.999 4.001 4.01 4.1
f(x): .20408 .2004 .20004 .19996 .1996 .19608
The limit would be x=.2
**Alright now for something I didn’t understand. Something that I thought was kind of confusing was infinite limits. The ones that I’m having trouble with are probably really simple, but I don’t get exercise 1-4 on 1.5. I don’t understand how you find if x approaches infinity or –infinity. Also, I don’t know what you’re supposed to plug in for problems 37 and 39. (When it’s like numbers from the right or left) If anyone could help me, I would appreciate it.
1st Blog
What's up, this is Chad.
Wellllll we are in calculus now and for the first 2 weeks we have been relearning what we learned at the end of advanced math. We have relearned everything about limits. We have learned continuity and numerical limits. Sooo i guess im going to talk about these things now.
CONTINUITY
Continuity basically means where does the graph exist and stuff like that. There are a couple things that make a graph not exist in some areas. These things are removables, verticle asymptotes, jumps, etc. I cant put a picture of a graph on here but i will describe these things the best i can. A removable is basically a hole in the graph. It should look like a little unfilled in hole. A vertical asymptote would look like two lines going up or going down at a certain number. These can do one of two things. They can either both be going the same direction or they can be going two different directions (the directions being either up or down). If these two match going one way then it can equal either infinite or negative infinite. Of course infinite means they both go up and negative infinite means they both go down. Last but not least there is a jump. A jump is where there are two graphs and when they get to a certain number they both stop. So lets say there is a graph of five. When the graph of five stops attttt 2, a graph of -3 may start the other way. That is probably really hard to understand but its the best i can do.
NUMERICAL LIMITS
Well a numerical limit is basically when you need to use the chart that we learned. The chart should subtract (from left to right on the chart) .1 , .01, .001 , then it should add .001, .01, .1 . You should add and subtract these numbers from the number you are using to plug into the limit.
Lets say you have a limit at x approaches 3 of x^2+2x+4. Now plug into the graph. The graph should come out saying (from left to right)
18.21 18.92 18.992 x 19.008 19.08 19.81
from the left going to the middle these numbers are approaching 19. Now from the right approaching left they are also approaching 19. So the limit of this equation is 19. You dont always have to use the graph. You only use it if you plug in and get undefined. I showed you that as an example because it was easy. Anyways, thats all i really have to say for now. So good luck in calculus ap everyone. I know ill need it.
Saturday, August 28, 2010
Mary Graci - Blog #1
Hey, this is Mary Graci. We’ve finally started the new year and now we’re in Calculus. To summarize these first two weeks, we’ve pretty much just reviewed what we learned at the end of the year in Adv. Math last year. But anyway, that review was of limits. We first re-learned how to create tables when the limit has a zero in the denominator. An example from our homework:
Ex. 1) lim as x approaches 2 ((x-2)/(x^2-4))
First, created a table by subtracting .1, .01, and .001 from 2, and adding .001, .01, and .1 to 2. Then plug those numbers into the equation. In this case, the answers are 100, 10000, 1000000, and 1000000, 10000, 100. Now compare both sides of the table and determine what each side is approaching, and obviously both sides are approaching infinity. So, from this you get infinity as your limit.
Next, we learned how to solve limits numerically without a chart, when there isn’t a zero in the demoninator. (It also happens to be the easiest way to solve limits!) All you have to do is plug in the x value given into the equation and, as easy as can be, you get the limit! But, an example anyway:
Ex. 2) lim as x approaches -3 (3x+2) = -7
Another problem may involve a trig function, but essentially, it works the same way.
Ex. 3) lim as x approaches 1 (cos pi x /3) = cos pi/3 = ½
**hint: don’t forget your trig chart!!!
There are, however, two special trig limits:
1) lim as x approaches 0 (sin x/x) = 1
2) lim as x approaches 0 (1-cos x/x) = 0
Besides solving limits numerically, you can also use algebra (factoring out discontinuities to make the problem simpler and then plugging in the x value to get the limit) and graphing (plugging the equation into your calculator, looking at the graph, and figuring out the limit that way.)
Besides that, I basically understood everything this week. It was all pretty easy and I did pretty good on the quizzes (especially the one Friday!) I don’t really have any questions this week, but I know it’s just going to get harder from here on out.
Ex. 1) lim as x approaches 2 ((x-2)/(x^2-4))
First, created a table by subtracting .1, .01, and .001 from 2, and adding .001, .01, and .1 to 2. Then plug those numbers into the equation. In this case, the answers are 100, 10000, 1000000, and 1000000, 10000, 100. Now compare both sides of the table and determine what each side is approaching, and obviously both sides are approaching infinity. So, from this you get infinity as your limit.
Next, we learned how to solve limits numerically without a chart, when there isn’t a zero in the demoninator. (It also happens to be the easiest way to solve limits!) All you have to do is plug in the x value given into the equation and, as easy as can be, you get the limit! But, an example anyway:
Ex. 2) lim as x approaches -3 (3x+2) = -7
Another problem may involve a trig function, but essentially, it works the same way.
Ex. 3) lim as x approaches 1 (cos pi x /3) = cos pi/3 = ½
**hint: don’t forget your trig chart!!!
There are, however, two special trig limits:
1) lim as x approaches 0 (sin x/x) = 1
2) lim as x approaches 0 (1-cos x/x) = 0
Besides solving limits numerically, you can also use algebra (factoring out discontinuities to make the problem simpler and then plugging in the x value to get the limit) and graphing (plugging the equation into your calculator, looking at the graph, and figuring out the limit that way.)
Besides that, I basically understood everything this week. It was all pretty easy and I did pretty good on the quizzes (especially the one Friday!) I don’t really have any questions this week, but I know it’s just going to get harder from here on out.
8/28/10
Well this week was basically a review of what we learned at the end of the year last year in pre-calc. Last week we went over limits and the different ways you can find them. For instance, the first thing you can do if you are asked to find the limit of a function is plug in the number they give to you as x. That works only when you get a number after you solve it. If you get zero as the denominator, then you have to try a new method. Another way to find a limit is to factor. If you’re asked to find the limit of a function that is a fraction, like for instance (x+2)/(x^2+3x+2), you can factor and cancel and then plug in the number they give you as x. Another way to find a limit (also if you get zero as the denominator first) you can use the table in your calculator and plug in the correct values. This week we learned two more special ways to find limits.—rationalizing and the Squeeze Theorem, and how to discuss the continuity of functions. I thought rationalizing was really easy, so here’s an example problem where you’d have to use that to find the limit:
Ex. 1) Find the limit (if it exists)… x>3lim ((sqrt of (x+1) – 2) / (x-3)*sqrt=square root/x>3=x goes to 3
*All “rationalizing” really means is to simplify the problem more so that it’s easier to solve. You can use rationalizing when you have a fraction, a square root, and something being added or subtracted.
*So for this problem you can use rationalizing since you have a square root and something being subtracted in the numerator. So the first thing you do is multiply the numerator and denominator of the problem by the numerator’s reciprocal.>> (sqrt of x+1) + 2
*Then multiplying the top out, you get x+1-4
*And for the denominator you get (x-3)((sqrt of x+1) +2)
*Simplifying the top you get (x-3)…So you can cancel out the (x-3)’s on the top and bottom of the fraction. Now you should be left with >>> (1)/((sqrt of x+1)+2)
*Now you can plug in 3 into that equation to get the limit.
So you get, (1)/(2+2) which is 1/4
**I basically understood everything this week. The only thing I’m a little iffy with is the Intermediate Value Theorem. I looked at the example in my notes, but I’m still not sure how I would do a problem like that on my own. The part where you would have to look at the graph confuses me.
Ex. 1) Find the limit (if it exists)… x>3lim ((sqrt of (x+1) – 2) / (x-3)*sqrt=square root/x>3=x goes to 3
*All “rationalizing” really means is to simplify the problem more so that it’s easier to solve. You can use rationalizing when you have a fraction, a square root, and something being added or subtracted.
*So for this problem you can use rationalizing since you have a square root and something being subtracted in the numerator. So the first thing you do is multiply the numerator and denominator of the problem by the numerator’s reciprocal.>> (sqrt of x+1) + 2
*Then multiplying the top out, you get x+1-4
*And for the denominator you get (x-3)((sqrt of x+1) +2)
*Simplifying the top you get (x-3)…So you can cancel out the (x-3)’s on the top and bottom of the fraction. Now you should be left with >>> (1)/((sqrt of x+1)+2)
*Now you can plug in 3 into that equation to get the limit.
So you get, (1)/(2+2) which is 1/4
**I basically understood everything this week. The only thing I’m a little iffy with is the Intermediate Value Theorem. I looked at the example in my notes, but I’m still not sure how I would do a problem like that on my own. The part where you would have to look at the graph confuses me.
Taylor Rodriguez Blog #1
Hello! Its Taylor Rodriguez.
This is my first blog of the year so I suppose the first blog of the year calls for a summary of the first chapter of the year.
So this summary will be on estimating limits numerically.
First Ill give some helpful hints that you need to remember when estimating limits numerically
• Remember that estimating limits numerically means to set up and complete the table
The first step to setting up the table is to step up the left side of the graph. To set up the left side of the graph you subtract .1 .01 .001 from the number that X is approaching in the problem and set the subtracted answers to each number into three boxes.
The next box will simply have the number x is approaching in it
The final three boxes will have the results of adding .001 .01 .1 to the number x is approaching respectively.
The first step to estimating the limit is complete
• To complete the table you must enter the given equation into your y=
• After you will hit “2nd” and table and enter in the numbers that you solved for in the seven boxes of the first step.
• now you will fill in the bottom seven boxes of the chart with the results from the table
• (( a quick way to know if you’ve done the steps correctly is to make sure that the box under the box with the number x is approaching should have an error on the table.
Finally to estimate the limit you will read each side toward the number x is approaching and decipher what number each side is headed toward as it approaches the given number x is approaching.
• Remember the three boxes on the left side will read toward the right and the right three boxes will read toward the left
You have the possibility to have two different out comes
• If the numbers on each side of the table match then the number they are headed toward is the estimated limit
• If the numbers on each side of the table do not match the limit does not exist.
I will now give an example of each outcome
EX:1
Lim X-> 0
F(x)= x/ squareroot of (x + 1) -1
The top half of the table will read
[-.1] [-.01] [-.001] [0] [.001] [.01] [.1]
The bottom half of the table will read
[1.9487] [1.995] [1.995] [ERROR] [2.0005] [2.005] [2.0488]
From the left the numbers are approaching 2
From the right the numbers are also approaching 2
Therefore the estimated limit is 2
Ex:2
Lim X-> 0
F(x)= sin 1/x
The top half of the table will read
[2/pi] [2/3pi] [2/5pi] [0] [2/7pi] [2/9pi] [2/11pi]
The bottom half of the table will read
[1] [-1] [1] [ERROR] [-1] [1] [-1]
From the left the numbers are not approaching anything
From the right the numbers are also not approaching anything
Therefore the estimated limit DNE
This is my first blog of the year so I suppose the first blog of the year calls for a summary of the first chapter of the year.
So this summary will be on estimating limits numerically.
First Ill give some helpful hints that you need to remember when estimating limits numerically
• Remember that estimating limits numerically means to set up and complete the table
The first step to setting up the table is to step up the left side of the graph. To set up the left side of the graph you subtract .1 .01 .001 from the number that X is approaching in the problem and set the subtracted answers to each number into three boxes.
The next box will simply have the number x is approaching in it
The final three boxes will have the results of adding .001 .01 .1 to the number x is approaching respectively.
The first step to estimating the limit is complete
• To complete the table you must enter the given equation into your y=
• After you will hit “2nd” and table and enter in the numbers that you solved for in the seven boxes of the first step.
• now you will fill in the bottom seven boxes of the chart with the results from the table
• (( a quick way to know if you’ve done the steps correctly is to make sure that the box under the box with the number x is approaching should have an error on the table.
Finally to estimate the limit you will read each side toward the number x is approaching and decipher what number each side is headed toward as it approaches the given number x is approaching.
• Remember the three boxes on the left side will read toward the right and the right three boxes will read toward the left
You have the possibility to have two different out comes
• If the numbers on each side of the table match then the number they are headed toward is the estimated limit
• If the numbers on each side of the table do not match the limit does not exist.
I will now give an example of each outcome
EX:1
Lim X-> 0
F(x)= x/ squareroot of (x + 1) -1
The top half of the table will read
[-.1] [-.01] [-.001] [0] [.001] [.01] [.1]
The bottom half of the table will read
[1.9487] [1.995] [1.995] [ERROR] [2.0005] [2.005] [2.0488]
From the left the numbers are approaching 2
From the right the numbers are also approaching 2
Therefore the estimated limit is 2
Ex:2
Lim X-> 0
F(x)= sin 1/x
The top half of the table will read
[2/pi] [2/3pi] [2/5pi] [0] [2/7pi] [2/9pi] [2/11pi]
The bottom half of the table will read
[1] [-1] [1] [ERROR] [-1] [1] [-1]
From the left the numbers are not approaching anything
From the right the numbers are also not approaching anything
Therefore the estimated limit DNE
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