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Questions on calc 1

  1. Jun 30, 2012 #1
    Instead of making many threads on different questions, I have decided to make one thread and ask a variety of questions here instead.

    1. The problem statement, all variables and given/known data

    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.


    3. The attempt at a solution

    Is the answer this, as limx[itex]\rightarrow-3^+[/itex]g(t)= t+4/t+3
     
    Last edited: Jun 30, 2012
  2. jcsd
  3. Jun 30, 2012 #2

    SammyS

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    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.
     
  4. Jun 30, 2012 #3

    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)...
     
  5. Jun 30, 2012 #4

    HallsofIvy

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    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.

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

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

    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.

    Ah. So the difficulty is English. That's to be expected on a forum on the "World Wide Web". What is your native language?


    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)".
     
  6. Jun 30, 2012 #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.
     
  7. Jun 30, 2012 #6

    SammyS

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    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.
     
  8. Jul 1, 2012 #7
    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.
     
  9. Jul 1, 2012 #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?
     
  10. Jul 1, 2012 #9

    eumyang

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    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).
     
  11. Jul 2, 2012 #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]
     
  12. Jul 2, 2012 #11

    Yes, I do not how to find the vertical asymptote of the function.
     
  13. Jul 2, 2012 #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).
     
  14. Feb 6, 2014 #13
    Wow.... calc one was so long ago.
     
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