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Calculus Help with Limits (check my work please)

  1. Feb 10, 2008 #1
    [SOLVED] Calculus Help with Limits (check my work please)

    Some problems, I just couldn't do on my own, but I've done some problems with answers provided here. Please check my work and show me how to do the problems I couldn't do. I would appreciate it.


    Problem 1:

    Lim
    x--> 3^- (left side)
    x--> 3^+ (right side)
    x--> 3

    Expression:

    |x+3| / (x-3) <== absolute value expression

    --------------------------

    Problem 2:

    lim x--> 3^-(left side)

    f(x) , where

    5x-7 if x is less than or equal to 3
    and
    x^2+1 if x is greater than 3

    Answer: f(x) = 8 (I'm not sure if it is right. Correct me if I'm wrong)

    ---------------------------

    Problem 3:

    lim
    x-->3

    (x^2-2x-3)/(x^2-6x+9)

    Answer: 1 (I'm not sure if it is right. Correct me if I'm wrong)

    ---------------------------

    Problem 4:

    lim
    x--> 2^- (left side)
    x--> 2^+ (right side)

    (3x+1)/(x-2)

    --------------------------

    Problem 5:

    lim
    x-->3

    (2x+6)/(x)

    Answer: 4 (I couldn't figure it out whether it was 4 or 8.)
     
  2. jcsd
  3. Feb 10, 2008 #2

    HallsofIvy

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    What happens if x is close to 3? Suppose x= 3.001 or x= 2.999. What is |x+3|/(x-3) in each of those cases? What happens if x is even closer to 3?
    If x< 3 which is what "3-" means, then f(x)= 5x-7. That's linear so continuous so the limit is just f(3)= 5(3)- 7= 8.

    That's a rational function so it's continuous wherever the denominator is not 0. Does that happen here? if x= 3, it is (32- 2(3)-3)/(32- 6(3)+ 9)= (9- 6- 3)/(9- 18+ 9)= 0/0. We can't just set x= 3 and find the limit. But the fact that both numerator and denominator are 0 at x= 3 tells us that x- 3 is a factor of each:
    (x2- 2x- 3)/(x2- 6x+ 9)= (x- 3)(x+ 1)/(x- 3)(x- 3). As long as x is not 3, we can cancel the two (x-3) factors: (x+1)/(x- 3). Now what is the limit of that as x goes to 3?

    Again, what happens if x is very close to 2? What if x= 2.0001? What if x= 1.999?
    Notice that in both this and problem 1, the denominator is 0 at the "target" point but the numerator is not

    What is that fraction equal to when x= 3? Why would you even think "8"?
     
  4. Feb 11, 2008 #3
    Will problem 4 be x = negative infinity and x = negative infinity?
     
  5. Feb 11, 2008 #4
    hello?
     
  6. Feb 11, 2008 #5

    Hootenanny

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    Close, but no cigar.
    Hi there :tongue2:
     
  7. Feb 11, 2008 #6

    HallsofIvy

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    No, problem 4 tells you that x will be close to 3. It doesn't ask for values of x!
     
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