2 calculus problems

Discussion in 'Community Discussion' started by macman2790, Jan 27, 2007.

  1. macman2790 macrumors 6502a

    macman2790

    Joined:
    Sep 4, 2006
    Location:
    Texas
    #1
    These two problems have been bugging me for a while now. on problem 1, you have to use limit laws, not by using numbers near 7. the answer is -1/16.

    on problem 2, i just couldnt figure it out, but i do understand the greatest integer function to a certain extent, i even used a calculator that had it on there. thanks in advance.

    1.Evaluate the limit if it exists:
    lim (square root of(x+2) - 3) divided by x-7
    x approaches 7

    2. This one uses the greatest integer function:

    if f(x) = {x} + - {x} show that lim x approaches 2 f(x) exists but is not equal to f(2).
     
  2. Rodimus Prime macrumors G4

    Rodimus Prime

    Joined:
    Oct 9, 2006
    #2
    this is not cal I is it.

    I has been a few years since I have done calculus. 2.5 years I believe to be exact and nearly 3 years since cal II so I am a little rusty

    Something tells me that the first one is not (neg) infinity when come from the negative (from 0 to 7) and positive infinity when coming from the positive side (10 to 7) because I believe that one of the laws is because the top power of x is less than the bottom it one of those things that prevents it but as I said I am really out of practice and those are rules I never really understood I just memorized and so I really really rusty on them.

    the 2nd part well I have no clue and it would require me to take out my calculus book and it 3:30am here. And I not even sure I could find the answer then so there is not a chance of me even looking at it until tomorrow any how.

    Best of luck 2 you.
     
  3. theyoda3 macrumors member

    Joined:
    Sep 27, 2006
    #3
    Ok I can help you with the first one. Use L'Hospital's Rule to get the limit. So take the derivative of the top of the fraction, then take the derivative of the bottom of the fraction. Then take the limit of the new fraction. Sub in 7 for x, and that is the limit as x approaches 7. The answer is really 1/6. I don't know how you got -1/16.

    Just as a note, it is because the limit as x approaches from the right and from the left is the same that allows you to just plug in 7 for x using L'Hospital's Rule.

    Haha, I like how we replied at the same time.

    Can you please clarify #2, is that a ± sign? Is it f(x) = {x} + - {x} , as x approaches 2*f(x) exists, but is not equal to f(2) = {2} + - {2} ?
     
  4. Rodimus Prime macrumors G4

    Rodimus Prime

    Joined:
    Oct 9, 2006
    #4
    lol I can not beleive I forgot L'Hospital's Rule as many times as I used that damn thing and it saved my sorry ass. But as I said it been a long time since I have done calculus. And just about 3 years since I have even used L'Hospital's Rule since we never used it in Cal III or the 2 other upper levels maths I took after that.
     
  5. macman2790 thread starter macrumors 6502a

    macman2790

    Joined:
    Sep 4, 2006
    Location:
    Texas
    #5
    the funny thing is this is Cal I and we haven't even discussed derivatives yet.
    #2 is shoud say + {-x) i mean instead of what it says. and yes your right about 1/6, -1/16 was a different one.
     
  6. theyoda3 macrumors member

    Joined:
    Sep 27, 2006
    #6
    Ok here is #2. The largest integer function is defined by {x}= the largest integer that is less than or equal to x. So say you have {2.5}, the largest integer less than or equal to 2.5 is 2 because 3 is bigger. Thus if 2≤x<3, then the integer is 2. For {-2.5}, the largest integer less than or equal to -2.5 is -3 because -2 is actually bigger than -3.

    Now that that is understood lets look at the problem. So the limit of {x} + {-x} as x approaches 2*f(x) is going to be -1. This is because for every integer there is only one point that is inconsistent with the rest. You will get a line of y=-1 at all points except where x= an integer. So lets say you are looking at a small portion of the graph between x=2 and x=3. At 2, {2}=2 and {-2}=-2; at 2.5, {2.5}=2 and {-2.5}=-3; and at 3. {3}=3 and {-3}=-3. So at x=2 and x=3 the function equals 0, but at ±2.5 the function equals -1. If you look at more numbers you will see that 2<x<3 in the function f(x) will equal -1. This also works with the rest of the numbers. Any integer<x<another integer for the function f(x) will equal -1. This means it's limit as it approaches 2*f(x) = -1 because the limit of the entire function is -1.

    Now for the second part. f(2) = {2} + {-2}. If you try this you will see that {2} = 2 and {-2} =-2. Therefore f(2)=0.

    As a result, the limit of {x} + {-x} as x approaches 2*f(x) = -1 and is not equal to f(2)=0. -1≠0.

    That's the best explanation I can muster up right now.

    -Yoda
     
  7. theyoda3 macrumors member

    Joined:
    Sep 27, 2006
    #7
    Ok, then you need to use the Quotient Rule. I don't remember exactly why it works so look it up in your book. I can give you instructions that lead to the correct answer though.

    First multiply by the conjugate of the numerator to simplify it. [(sqrt(x+2)-3) / (x-7)] * [(sqrt(x+2)+3)/(sqrt(x+2)+3)]. You will get [1/(sqrt(x+2)+3)]. This lets you set LIM [(sqrt(x+2)-3) / (x-7)] = LIM [1/(sqrt(x+2)+3)]. Using the Quotient Rule you can move the limit inside the square root to get [1/(LIM sqrt(x+2)+3)]. Now you evaluate the LIM sqrt(x+2) as x approaches 7. You will get the limit = 9. Plug 9 back into the previous fraction to get [1/(sqrt(9)+3)]. This is equal to [1/(3+3)], which leads to the final answer of 1/6.

    -Yoda
     
  8. ®îçhå®? macrumors 68000

    ®îçhå®?

    Joined:
    Mar 7, 2006
    #8
    The Quotient Rule is dy/dx = ((v.du/dx)-(u.dv/dx))/(v2)
     
  9. theyoda3 macrumors member

    Joined:
    Sep 27, 2006
    #9
    It's not that kind of quotient rule. It is a Quotient Rule of limits not derivatives. It is the Quotient Rule that says the lim of x/y = lim x / lim y.

    Looking back at the problem it is actually 2 rules: the Addition Limit Rule and the Quotient Limit rule. The Addition Limit Rules says lim (x + y) = lim x + lim y.

    So starting at LIM [1/(sqrt(x+2)+3)] if you use these two rules it becomes [LIM 1/(LIM sqrt(x+2)+LIM 3)]. So LIM 1=1 , LIM 3=3, and LIM sqrt(x+2)= 9. These are all limits for when x approaches 7. You must maintain the x approaches 7 part of the limit evaluation throughout the entire problem.

    -Yoda
     
  10. macman2790 thread starter macrumors 6502a

    macman2790

    Joined:
    Sep 4, 2006
    Location:
    Texas
    #10
    thanks, i'll see how this works
     

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