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Your conclusion is correct, but the same comments I made on your other piecewise function thread also apply here so I suggest you read that first (it's the same kind of set-up, just with two pieces rather than three).
The question did ask for you to use a graph. So what do you think the graph looks like and in particular what do you think it looks like near x = -1?
Read my reply to the other thread. I am now learning how to use this site. I first have to learn how to upload picture attachments before we can talk about graphs of functions. You are right in your comment. However, I want to take it easy as I make my way through the Ron Larson textbook. No need to rush. No need to panic. No exam to prepare for. Get it?
Your conclusion is correct -- however, there are a few technical points to note.
nycmathguy wrote:Find the limit of x^2 as x tends to 1 from the left side.
What you really want to be saying is that you are finding the limit of f(x) as x tends to 1 from the left, rather than the limit of x^2.
nycmathguy wrote:(1)^2 = 1
Strictly speaking, this is how you evaluate the limit of f(x) as x approaches 1 from the left, but this isn't how you evaluate the left-hand limit of x^2, this is just substituting in x = 1 -- although you can get away with it here. (This is because x^2, 2 and -3x + 2 are all continuous functions, so the left and right-hand limits for each function exist, are equal, and are equivalent to their usual limits.) The way you ought to be thinking about it is:
It'll lead to the same answer, but conceptually this is along the sort of lines you'll be required to think through when you look at the epsilon-delta type questions.However, the question did ask for you to use a graph. So what do you think f(x) might look like on a graph? What kind of shape does it have as you move from left to right? And in particular what happens near x = 1?
1. Let's not get too technical here.
2. I am not thinking about graphing piecewise functions at this early stage in my self-study of calculus.
3. As far as the delta/epsilon definition of a limit, I may or may not include this in my self-study of this course.
4. I am not a classroom student. My college days ended in 1994.
5. Why did you feel a need to bring delta into this reply? I don't know what you did here. I have not taken calculus in a classroom setting and at my age, I never will.
Use a graph to investigate limit of f(x) as x tends to c at the number c.
Note: f(x) is a piecewise function.
P. S. I tackled the problem algebraically not by way of graph.
When I learn how to post images, I will use more graphs throughout my study of calculus.
Top portion of piecewise function: x^2 if x < 1
Middle portion of piecewise function: 2 if x = 1
Bottom portion of piecewise function: -3x + 2 if x > 1
at c = 1
Find the limit of x^2 as x tends to 1 from the left side.
(1)^2 = 1
Find the limit of -3x + 2 as x tends to 1 from the right side.
-3(1) + 2 =
-3 + 2 = -1
LHL DOES NOT = RHL.
I say the limit of f(x) does not exist.
NOTE: I will use DNE to mean Does Not Exist moving forward in my study of limits as famously done in all calculus textbooks.
Use a graph to investigate limit of f(x) as x tends to c at the number c.
Note: f(x) is a piecewise function.
Note: I did not use a graph of the given function.
Top portion of piecewise function: x^3 if x < -1
Bottom portion of piecewise function: x^2 - 1 if x > -1
at c = -1
Solution:
Find the limit of x^3 as x tends to -1 from the left.
(-1)^3 = -1
Find the limit of x^2 - 1 as x tends to -1 from the right.
(-1)^2 - 1
1 - 1 = 0
The LHL DOES NOT = RHL.
Thus, the limit of f(x) does not exist.
That's what I said in post 2.
Bob
I like calculus already. I will post a few more piecewise functions before moving on.
I do hope my questions and replies help others as well.
As the function has 3 parts I would have included that y = 4 there too, but I'm probably being very picky. What you have said is excellent.
Picture step 2. Do you have a picture that you want to appear on the forum using imgur?
On my laptop I can do a screen shot and then edit and save the resulting image. Let's try any image to start with. We can look at how you edit a shot later.
The app should have an option to upload the image to your account. Let me know if you have been able to do that.
Bob
I took a picture of the piecewise function for this thread. Image has been posted to the app. What's next?
As the function has 3 parts I would have included that y = 4 there too, but I'm probably being very picky. What you have said is excellent.
Picture step 2. Do you have a picture that you want to appear on the forum using imgur?
On my laptop I can do a screen shot and then edit and save the resulting image. Let's try any image to start with. We can look at how you edit a shot later.
The app should have an option to upload the image to your account. Let me know if you have been able to do that.
Bob
I don't have any particular picture in mind to upload. However, I will take a picture of this piecewise function as step 2.
Good. I've given the next step on your other post. So let's call this thread finished.
Bob
What other post?
Use a graph to investigate the limit of f(x) as x tends to c at the given c number.
Let c = 2
Top portion of piecewice function: x + 2 if x < 2
Middle portion of piecewise function: 4 if x = 2
Bottom portion of piecewise function: x^2 if x > 2
Note: f(x) is a three-part function.
Solution:
Find the limit of (x + 2) as x tends to 2 from the left side.
(2 + 2) = 4
Find the limit of x^2 as x tends to 2 from the right side.
(2)^2 = 4
LHL = RHL
Thus, the limit of f(x) as x tends to 2 is 4.
Is this right?
Note: Sorry that I could not upload a picture for clarity. Still waiting for step 2 for uploading pictures and graphs.
P. S. I don't have a computer or laptop. All this typing work is done on my cell phone. This is true.
hi nycmathguy
Funnily enough there's another member in NYC who is studying exactly the same topics as you who also asked exactly this question.
He wanted to be told the steps just one step at a time. I'm guessing you would like the same.
The first step is to sign up to www.imgur.com.
Post back when you have done that and I'll tell you step 2.
Bob
I signed up at www.imgur.com.
I had to download the app.
What is next?
Bob wrote:OK. I'll say what I think is the answer.
I think that all values of c need to be considered.
(1) c< 0 lim = 2c+1
(2) c> 0 lim = 2c
(c) c=0, indeterminate.
That's the best I can offer I'm afraid.
Bob
Unfortunately, this is an even number problem in the book to which there is no answer in the back section. I don't know if you are right or wrong.
I went back to see if the author has a sample problem for piecewise functions.
Luckily, Sullivan does provide a sample for a similar question. I will use the sample question in the textbook as a guide to help me solve problem 30.
30. Investigate the limit of f(x) as x tends to c at the number c.
The number c is given to be 0.
{2x + 1 if x ≤ 0....top portion of piecewise function.
{2x if x > 0..........bottom portion of piecewise function.
We want to do find the limit as x tends to 0 from the left and right.
Find the limit (2x + 1) as x tends to 0 from the left.
2(0) + 1 = 1
Find the limit of 2x as x tends to 0 from the right.
2(0) = 0
The left handed limit DOES NOT EQUAL the right handed limit.
Thus, the limit of f(x) as x tends 0 does not exist.
You say?
OK. I'll say what I think is the answer.
I think that all values of c need to be considered.
(1) c< 0 lim = 2c+1
(2) c> 0 lim = 2c
(c) c=0, indeterminate.
That's the best I can offer I'm afraid.
Bob
Unfortunately, this is an even number problem in the book to which there is no answer in the back section. I don't know if you are right or wrong.
I'll assume you don't want help with posting pictures of graphs then.
B
All I need is the steps. This is my first time joining a site that does not have an attachment for pictures.
hi nycmathguy
Funnily enough there's another member in NYC who is studying exactly the same topics as you who also asked exactly this question.
He wanted to be told the steps just one step at a time. I'm guessing you would like the same.
The first step is to sign up to www.imgur.com.
Post back when you have done that and I'll tell you step 2.
Bob
There are over 20 million people in NYC. Why are you startled? Are you saying I am the other guy? What's his username? NYC is divided into 5 boroughs. Where does he live? Am I the only person studying calculus in the Big Apple?
Is there a bit missing from this question? The definitions for the function do not have any 'c' in them, so how are we supposed to know what c is. Or has your finger just slipped and typed c when you meant x?
I've sketched the graph and there's a discontinuity at x = 0. The left limit is y=1 and the right limit is y = 0.
So the usual conclusion is to say there is no consistent limit at x = 0.
Bob
Hello Bob. Good morning. I did not make a typo.
Here are the specific instructions:
Use a graph to investigate the limit of f(x) as x tends to c at the number c.
Sullivan provided the value of c to be 0. I decided to solve it algebraically. However, if I must check the limit from the left and right, how is this done algebraically in terms of piecewise functions?
Investigate the limit of f(x) as x tends to c at the number c.
{2x + 1 if x ≤ 0....top portion of piecewise function.
{2x if x > 0..........bottom portion of piecewise function.
The number c = 0.
Solution:
If c = 0, I should use the top portion of f(x).
That is, 2x + 1.
Let x = 0.
2(0) + 1 = 0 + 1 = 1.
I say the limit of f(x) is 0 when c = 0.
I hope this is correct.
hi nycmathguy
Welcome to the forum.
I've just logged in and see you have two posts on limits. As this one came first I'll start by replying here.
If you're starting out on calculus I wouldn't get too hung up on the definition. Despite what Stephen Hawking said, the universe hasn't got built in mathematical laws. It's just what we (ie. centuries worth of mathematicians) have made up to suit our purposes. I think the process which is usually described as differentiation from first principles is more important to grasp. If you follow that then the world of calculus will open up to you.
If you have a straight line with points (3,7) and (5,15) lying on it, then we can calculate the gradient thus;
The theory surrounding y = mx + c shows us that this answer holds good if you take other points on the line.
But what if the function leads to a curve when you graph it. A function like y = x^2 has a gradient that is continuously changing. So what to do if we want the gradient at (say) (2,4)
You can take a point close to (2,4), let's say (2.5,6.25) and calculate the gradient of the chord that joins the points.
A rough sketch of the curve will show you that this answer is too big for the gradient at (2,4)
If you take a point to the left such as (1.5, 2.25) we can do the calculation again:
Our sketch shows us that this answer is too small for the gradient at (2,4).
So we already know that the answer we want is between 3.5 and 4.5.
You can move closer, and try (let's say) (2.1, 4.41):
and closer still:
It would be easy to just move our second point on top of the first:
o divided by o could be anything; what number do you have to multiply 0 by to get 0?
So that doesn't help. But what we can observe is that our answers (4.5, 4.1, 4.01) are getting closer and closer to 4. That's what we mean by a limit. Is it true that the answers will keep getting closer and closer to 4.
You could try (2.000001, whatever that is when squared). Or use algebra:Let's call the second point (x + Δ, x^2+ 2x.Δx + Δx^2) and the first (x,x^2)
Δx means a little bit in the x direction and it's a single symbol not two symbols multiplied together.
Now I'm going to let Δx get smaller and smaller. We say it 'tends' to zero. But if I do that with the above expression I'll get 0/0 again. The 'trick' is to do some simplification first.
Now we can see why the answers with numbers got closer and closer to 4. That Δx means the answer is always above 4, but as it tends to zero, the gradient of the chord gets closer and closer to 2x.
That's the idea of a limit in practice. I did it without worrying about the formal definition. Much much later, if you get to University pure math courses the professors may start introducing the formal difinition just to make the subject 'properly rigorous', but you can get by without it. Newton invented the process (at the same time as Leibnitz) and I'm pretty sure they just developed the method without too much formality.
There's more here:
https://www.mathsisfun.com/calculus/introduction.html
Hope that helps,
Bob
I found a tutor on You Tube who explained the delta/epsilon method to prove that a limit does exist. He made it seem easy. I actually understood what he said. I may be
tempted to try a few examples in the coming chapters but not so sure right now.
Name: MySecretMathTutor on You Tube. Check him out. Tell me what you think.
Thank you.
Hope this satisfies all:
https://i.imgur.com/umbEqXL.gif
Bob
Beautiful graph. Nice colors. Tell me, how did you upload this graph? I may need to graph some functions in the coming chapters.
hi
Your working is all correct. You can differentiate the function (dy/dx = 2x) to get the gradient function and then substitute x = 1.
The normal to a curve at a point A is the line through A at right angles to the tangent.
So in your example, dy/dx = 2x (= 2 at x = 1). I think you already know that a line perpendicular to the tangent has gradient n
where nm = -1. So the gradient of the normal is -1/2. And it goes through (1,1) so you can use the same method as above to get the equation.If you know two points on a curve, A and B, the secant line is the line AB.
Bob
You said:
"So in your example, dy/dx = 2x (= 2 at x = 1). I think you already know that a line perpendicular to the tangent has gradient nwhere nm = -1. So the gradient of the normal is -1/2. And it goes through (1,1) so you can use the same method as above to get the equation."
Can you please provide a sample for this statement?
"If you know two points on a curve, A and B, the secant line is the line AB."
Can you please provide an example for this statement?
Can you please provide a graph using different color lines showing the tangent line, secant line and normal line all on the same xy-plane?
Thank you.
As I understand the introduction by Sullivan, there are 3 main objectives in
Calculus 1 aka First Semester Calculus:
1. Find the tangent line to a curve at a point (x, y).
2. Find the limit of functions.
3. Find the area under a curve between two points.
I also understand that Calculus 2 is mainly about integration. The author went on to say that most students greatly struggle in Calculus 2 due to the amount of integration techniques they must quickly learn.
Calculus 3 aka Multivariable Calculus mainly deals with 3D images. This course is also called Vector Calculus. The author went on to say that although Calculus 3 is no walk in the park, it is basic calculus compared to Advanced Calculus dealing with tensors.
I will be totally satisfied if I can "master" (see the quotation marks) the first-three calculus courses taken by most math, science and engineering students in colleges around the globe. I am not saying that tensor calculus is boring. I am simply saying that my interest does not extend beyond Calculus 3 as far as Calculus is concerned.
Differential Equations is a course that involves lots of integration. I may want to learn this material in the future. I DEFINITELY desire to learn Linear Algebra. I have seen videos clips of Linear Algebra and find its content super cool and enriching.
Questions:
1. In your opinion, why is Calculus 2 harder than 1 and/or 3?
2. Member Bob stated that as a beginning Calculus student, there is no need for me to study/learn the formal definition of a limit using delta and epsilon. I totally agree with Bob. What do you say?
3. Do you know Advanced Calculus aka Tensor Calculus? I saw a few minutes of this course on You Tube. Honestly, it is way beyond my knowledge of mathematics. It is also very confusing. No need for me to ever learn all this tensor stuff.
4. What is your favorite Calculus topic? Some people like limits, some like the derivative, still others like integration. How about you?
Find the equation of the tangent line to the curve y = x^2 at the point (1, 1). The slope of the tangent line is given to be 2.
The equation I need to find is the line that goes through or touches the curve at the point (1, 1) located in quadrant 1.
I am going to use the point-slope formula.
y - y_1 = m(x - x_1)
Let x_1 = 1
Let y_1 = 1
Let m = slope of the tangent line.
y - 1 = 2(x - 1)
y - 1 = 2x - 2
y = 2x - 2 +1
y = 2x - 1
The equation y = 2x - 1 is the tangent line at the point (1, 1).
Is any of this right?
Follow-up questions for the reader.
1. How do I find the slope of the secant line?
2. What about the slope of the normal line?
3. What is the BASIC DIFFERENCE between the slope of the tangent line, secant line and normal line?
4. In the question above, the slope of the tangent line is given. What if the slope is not given? How do I proceed?
Thank you.
Hi all;
I have simple problem of this concept that is
If there is sqrt 2 divided by sqrt 9
Then how we say (2÷9)whole sqrt
sqrt{2}/sqrt{9}
The sqrt{9} = 3.
Answer: sqrt{2}/3.
In decimal form, we get 0.4714045208.
Rounded to 4 decimal places, we get 0.4714.
Hi nycmathguy,
Welcome to the forum!
I think you are on the right path.
See the links for reference:
Thank you for the links. I look forward to learning calculus on this site as well as on You Tube and other places online. Look for my questions.
hi nycmathguy
Welcome to the forum.
I've just logged in and see you have two posts on limits. As this one came first I'll start by replying here.
If you're starting out on calculus I wouldn't get too hung up on the definition. Despite what Stephen Hawking said, the universe hasn't got built in mathematical laws. It's just what we (ie. centuries worth of mathematicians) have made up to suit our purposes. I think the process which is usually described as differentiation from first principles is more important to grasp. If you follow that then the world of calculus will open up to you.
If you have a straight line with points (3,7) and (5,15) lying on it, then we can calculate the gradient thus;
The theory surrounding y = mx + c shows us that this answer holds good if you take other points on the line.
But what if the function leads to a curve when you graph it. A function like y = x^2 has a gradient that is continuously changing. So what to do if we want the gradient at (say) (2,4)
You can take a point close to (2,4), let's say (2.5,6.25) and calculate the gradient of the chord that joins the points.
A rough sketch of the curve will show you that this answer is too big for the gradient at (2,4)
If you take a point to the left such as (1.5, 2.25) we can do the calculation again:
Our sketch shows us that this answer is too small for the gradient at (2,4).
So we already know that the answer we want is between 3.5 and 4.5.
You can move closer, and try (let's say) (2.1, 4.41):
and closer still:
It would be easy to just move our second point on top of the first:
o divided by o could be anything; what number do you have to multiply 0 by to get 0?
So that doesn't help. But what we can observe is that our answers (4.5, 4.1, 4.01) are getting closer and closer to 4. That's what we mean by a limit. Is it true that the answers will keep getting closer and closer to 4.
You could try (2.000001, whatever that is when squared). Or use algebra:Let's call the second point (x + Δ, x^2+ 2x.Δx + Δx^2) and the first (x,x^2)
Δx means a little bit in the x direction and it's a single symbol not two symbols multiplied together.
Now I'm going to let Δx get smaller and smaller. We say it 'tends' to zero. But if I do that with the above expression I'll get 0/0 again. The 'trick' is to do some simplification first.
Now we can see why the answers with numbers got closer and closer to 4. That Δx means the answer is always above 4, but as it tends to zero, the gradient of the chord gets closer and closer to 2x.
That's the idea of a limit in practice. I did it without worrying about the formal definition. Much much later, if you get to University pure math courses the professors may start introducing the formal difinition just to make the subject 'properly rigorous', but you can get by without it. Newton invented the process (at the same time as Leibnitz) and I'm pretty sure they just developed the method without too much formality.
There's more here:
https://www.mathsisfun.com/calculus/introduction.html
Hope that helps,
Bob
Hi Bob. I'm happy to be a new member. It's good to know that there's no need to worry about the formal definition of a limit using delta and epsilon. When I saw this delta/epsilon stuff in chapter 1, I quickly became discouraged.
I think finding the limit of functions via graphs, substitution and algebraically is better for me at this level of learning. So, I will do as you say and skip all the rigorous, pure math jargon. This is great news. I can actually learn to find limits without pure mathematics, abstract material.
Thank you for taking the time to type all this information. Thank you for the link. I will visit the link to read and study more about limits in addition to the Michael Sullivan textbook. I will return with questions I get stuck with. Math work will be shown. It is my hope to be a regular user for years to come.
Hello everyone. How are you? I want to learn calculus so badly. I plan to do a self-study through calculus l, ll, and lll. Before I think so far ahead, I need a clear, basic definition of the concept of a limit. Textbook language is never easy to grasp unless the student is gifted. I am not gifted mathematically but love the subject.
I understand the limit idea to be the following:
•The value a function is tending to without actually getting there.
•The height a function is trying to reach in terms of the y-axis.
•In terms of a point (x, y), the limit is (value the function is tending to, limit). In other words, x = the value a function is tending to and y = the limit, the height of the function.
Is my understanding of a limit clear or not?