Possible equality about lim x\rightarrow-3^+ g(t): g(-3) = (-3)+4/(-3)+3 = 1

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In summary, instead of creating multiple threads for different questions, the user has decided to ask various questions in one thread. The user's first question is about writing an equality for the limit of g(x) as x approaches -3 from the right. The user also asks for a definition of equality and if it is possible to write an equality for the given limit. The user is having trouble understanding the concept of equality and is using incorrect notation in their attempt at a solution. They apologize for any confusion caused by their misunderstanding of the notation.
  • #1
ME_student
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Instead of making many threads on different questions, I have decided to make one thread and ask a variety of questions here instead.

Homework Statement



Suppose that g(x)= x+4/x+3

Write an equality about lim x[itex]\rightarrow-3^+[/itex] g(t)

What is an equality?

Is it possible to write an equality about lim x[itex]\rightarrow-3^+[/itex] g(t), if so write it?

I can't answer this question because I don't know what an equality is.

The Attempt at a Solution



Is the answer this, as limx[itex]\rightarrow-3^+[/itex]g(t)= t+4/t+3
 
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  • #2
ME_student said:
Instead of making many threads on different questions, I have decided to make one thread and ask a variety of questions here instead.
That's not how things are done here at PF. I'm sure that the Moderators will not like that.

Homework Statement



Suppose that g(x)= x+4/x+3

Write an equality about lim x[itex]\rightarrow-3^+[/itex] g(t)

What is an equality?

Is it possible to write an equality about lim x[itex]\rightarrow-3^+[/itex] g(t), if so write it?

I can't answer this question because I don't know what an equality is.

The Attempt at a Solution



Is the answer this, as limx[itex]\rightarrow-3^+[/itex]g(t)= t+4/t+3
As for your question:

It's important to use parentheses where needed.

The expression you have written, t+4/t+3, literally means [itex]\displaystyle t+\frac{4}{t}+3\,.[/itex] I doubt that you meant that.

Write it as (t+4)/(t+3) if what you mean is [itex]\displaystyle \frac{t+4}{t+3}\,.[/itex]

What is an equality?
An equality is basically an equation; two items are equal. --- to be contrasted with an inequality; one item is greater than , or not less than another item.
 
  • #3
SammyS said:
That's not how things are done here at PF. I'm sure that the Moderators will not like that.
As for your question:

It's important to use parentheses where needed.

The expression you have written, t+4/t+3, literally means [itex]\displaystyle t+\frac{4}{t}+3\,.[/itex] I doubt that you meant that.

Write it as (t+4)/(t+3) if what you mean is [itex]\displaystyle \frac{t+4}{t+3}\,.[/itex]


An equality is basically an equation; two items are equal. --- to be contrasted with an inequality; one item is greater than , or not less than another item.


Okay thanks.

Why can't I use the same thread over and over again? I don't see a problem with it...
Lets wait and see what the mod says... And yes I meant to say (t+4)/(t+3)...
 
  • #4
ME_student said:
Instead of making many threads on different questions, I have decided to make one thread and ask a variety of questions here instead.
That makes it less likely that a question will be answered. Many people, when they have responded to a thread, don't look at it again.

Homework Statement



Suppose that g(x)= x+4/x+3
If you won't use LaTeX, at least use parentheses! Do you mean (x+4)/(x+3)?

Write an equality about lim x[itex]\rightarrow-3^+[/itex] g(t)
I have no idea what "3+" means. Do you mean [itex]x\rightarrow -3^{g(t)}[/itex]?

What is an equality?
Uh- its an equation.

[quoter]Is it possible to write an equality about lim x[itex]\rightarrow-3^+[/itex] g(t), if so write it?[/quote]
Well, I presume it is possible to write an equation about whatever the function is really supposed to be or they wouldn't have asked this question.

I can't answer this question because I don't know what an equality is.
Ah. So the difficulty is English. That's to be expected on a forum on the "World Wide Web". What is your native language?


The Attempt at a Solution



Is the answer this, as limx[itex]\rightarrow-3^+[/itex]g(t)= t+4/t+3

That doesn't make any sense. I still don't know what "[itex]3^+[/itex]" is supposed to mean. And you shouldn't have both "g(t)" and "(0t+4)/(t+3)".
 
  • #5
My native language is English, but I have a bit of a hard time with my vocabulary... :redface:

x[itex]\rightarrow3^+[/itex] means it's approaching 3 from the right side and x[itex]\rightarrow3^-[/itex] means it's approaching three from the left... You have never seen that before?

I almost named the title of the thread "Zach's calc 1 questions" so that people would understand that I will be asking multiple question within this thread, over and over again.
 
  • #6
ME_student said:
My native language is English, but I have a bit of a hard time with my vocabulary... :redface:

x[itex]\rightarrow3^+[/itex] means it's approaching 3 from the right side and x[itex]\rightarrow3^-[/itex] means it's approaching three from the left... You have never seen that before?

I almost named the title of the thread "Zach's calc 1 questions" so that people would understand that I will be asking multiple question within this thread, over and over again.
I'm quite sure that HallsofIvy has seen x→-3+, etc.

Your expression, limx[itex]\rightarrow-3^+[/itex]g(t)= t+4/t+3 , has several problems.

Assuming you mean [itex]\displaystyle \lim_{x\to\,-3^+}g(t)= \frac{t+4}{t+3}\,:[/itex]
Your limit has the variable, x, going to -3 from the right, but you have the variable, t, in your function expressions. (I see that you fixed that up in some lines of your Original Post, but not in others.)

Even if you replace t with x, throughout the expression, the statement, [itex]\displaystyle \lim_{x\to\,-3^+}g(x)= \frac{x+4}{x+3}\,,[/itex] makes no sense.
 
  • #7
SammyS said:
I'm quite sure that HallsofIvy has seen x→-3+, etc.

Your expression, limx[itex]\rightarrow-3^+[/itex]g(t)= t+4/t+3 , has several problems.

Assuming you mean [itex]\displaystyle \lim_{x\to\,-3^+}g(t)= \frac{t+4}{t+3}\,:[/itex]
Your limit has the variable, x, going to -3 from the right, but you have the variable, t, in your function expressions. (I see that you fixed that up in some lines of your Original Post, but not in others.)

Even if you replace t with x, throughout the expression, the statement, [itex]\displaystyle \lim_{x\to\,-3^+}g(x)= \frac{x+4}{x+3}\,,[/itex] makes no sense.

Then I must be misunderstanding the question. My apologies, I am still trying to familiarize myself with the forum as well.

EDIT: I meant as t approaches -3^+, not x approaches -3^+. For some reason I use multiple variables before I realize something is messed up... It can get very confusing.
 
  • #8
Problem 11.1 Suppose that g(t)=(t+4)/(t+3)

11.1.1 What is the vertical asymptote on the graph of y=g(t)?

11.1.2 Write an equality about lim t[itex]\rightarrow-3^+[/itex] g(t)?

11.1.3Write an equality about lim t[itex]\rightarrow-3^-[/itex] g(t)?

11.1.4 Is it possible to write an equality about lim t[itex]\rightarrow-3[/itex]g(t)? If so, write it?
 
  • #9
ME_student said:
11.1.1 What is the vertical asymptote on the graph of y=g(t)?
Do you know how to find the vertical asymptote?
11.1.2 Write an equality about lim t[itex]\rightarrow-3^+[/itex] g(t)?
11.1.3Write an equality about lim t[itex]\rightarrow-3^-[/itex] g(t)?
So what we are being asked are these:
[itex]\displaystyle \lim_{t\to\,-3^+}\frac{t+4}{t+3}\,[/itex] and
[itex]\displaystyle \lim_{t\to\,-3^{-}}\frac{t+4}{t+3}\,[/itex]
If this is your first exposure to limits, then maybe making a table of values (t vs. g(t)) would help. For the right-hand limit, choose the values -2.9, -2.99, -2.999, -2.9999 for t, and find the corresponding g(t) values. Then you can evaluate the right-hand limit. After that, make a new table for the left-hand limit, choosing new values for t (I'll let you figure those out).
 
  • #10
So when it comes to one sided limits, I generally try to do it algebraically rather than doing tables and what not.

Consider the following
[itex]\lim_{x\to a^{+}} f(x)=\lim_{n \to ∞}f(a+\frac{1}{n})[/itex]
and
[itex]\lim_{x\to a^{-}} f(x)=\lim_{n \to ∞}f(a-\frac{1}{n})[/itex]
 
  • #11
eumyang said:
Do you know how to find the vertical asymptote?

So what we are being asked are these:
[itex]\displaystyle \lim_{t\to\,-3^+}\frac{t+4}{t+3}\,[/itex] and
[itex]\displaystyle \lim_{t\to\,-3^{-}}\frac{t+4}{t+3}\,[/itex]
If this is your first exposure to limits, then maybe making a table of values (t vs. g(t)) would help. For the right-hand limit, choose the values -2.9, -2.99, -2.999, -2.9999 for t, and find the corresponding g(t) values. Then you can evaluate the right-hand limit. After that, make a new table for the left-hand limit, choosing new values for t (I'll let you figure those out).


Yes, I do not how to find the vertical asymptote of the function.
 
  • #12
This is what I got for 11.1.2

lim t[itex]\rightarrow-3^-[/itex] g(t)=-[itex]\infty[/itex]

and for lim t[itex]\rightarrow-3^+[/itex] g(t)=[itex]\infty[/itex]

For 11.1.4 I am still not sure... I don't think you can write an equality of lim t[itex]\rightarrow-3[/itex] g(t).
 
  • #13
Wow... calc one was so long ago.
 

FAQ: Possible equality about lim x\rightarrow-3^+ g(t): g(-3) = (-3)+4/(-3)+3 = 1

1. What is Calculus 1?

Calculus 1 is typically the first course in a university-level calculus sequence. It covers the fundamentals of differential and integral calculus, including limits, derivatives, and integrals.

2. What topics are typically covered in Calculus 1?

Calculus 1 typically covers topics such as limits, continuity, derivatives, optimization, and basic integration techniques. It may also include some applications of calculus, such as related rates and area/volume problems.

3. What skills or background knowledge do I need for Calculus 1?

Calculus 1 typically requires a strong foundation in algebra and trigonometry. It is also helpful to have a basic understanding of functions and their graphs.

4. How is Calculus 1 different from other math courses?

Calculus 1 is different from other math courses in that it focuses on the concepts of change and rates of change, rather than just solving equations. It also introduces the concept of limits, which is essential for understanding calculus.

5. How can I prepare for Calculus 1?

You can prepare for Calculus 1 by reviewing algebra and trigonometry concepts, as well as familiarizing yourself with the basics of functions and their graphs. You can also practice solving problems involving rates of change and limits to help build your understanding of calculus concepts.

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