Feedback on 05 Limits and Continuity

You will find here additional information about the various problems which students have asked about. Check here if you are having problems with specific exercises; you can also send e-mail to ken@ms.uky.edu.
  1. Corrections to the Homework web page:
    1. Question 2 had two identical answers -- but only one of them would have been graded as correct. It has been changed so there is a unique correct answer.
  2. (From bomarf 8/31/2000) Question 1:

    I thought that as x approched 3 that the limit wouldn't exist,

    The limit exists because as you approach x = 1 from either side, the values of the function get closer and closer to 3. Note that the limit does NOT depend on the value of the function at x = 3, but depends only of the values of the function for x near 3.

  3. (From beckerk 8/31/2000) Question 4:

    Why does the limit not exist? I thought it was approaching infinity in both directions.

    No, from the left, it approaches negative infinity; and from the right, it approaches plus infinity. So the two one sided limits are not equal and the limit does not exist.

  4. (From beckerk 8/31/2000) Question 4:

    i didn't understand what the question was asking for, it was unclear to me

    The problem is asking for the value of the limit as x approaches 2 of (x^3-8)/(x-2). The first thing to try is to see if the denominator has value 0 at x = 2. If not, then you can calculate the limit by just substituting the x = 2 into the fraction. Unfortunately, the denominator is zero; so it is harder. The second thing to try is to see if one can't factor the numerator and cancel. You can do this. One has x^3-8 = (x-2)(x^2+2x+4) and so the limit is the same as that of x^2+2x+4. Here we can just substitute x=2 and so the limit is 12.

  5. (From prevatth 8/30/2000) Question 6:

    Please work this problem out- I am unsure of how to do it. Thank you.

    First check to see if the limit of the denominator is zero -- it is; so you can't just substitute in x = 2 to get the limit. In this case, the limit of the denominator as x approaches 2 is 2^2 - 1 = 3. So the numerator is positive for all x near 2.

    1. For x near 2 but to the left of (i.e. smaller than) 2, x - 2 is negative and close to zero. So the value of the fraction will be negative and very large in absolute value for x very close to and less than 2. This means the limit as x approaches 2 from the left of (x^2-1)/(x-2) is negative infinity.
    2. For x near 2 but to the right of (i.e. larger than) 2, x - 2 is positive and close to zero. So the value of the fraction will be positive and very large in absolute value for x very close to and larger than 2. This means the limit as x approaches 2 from the left of (x^2-1)/(x-2) is positive infinity.
    Since the limit from left is not equal to the limit from the right, the limit does not exist.
  6. (From prevatth 8/30/2000) Question 7:

    Please work out this problem; I do not understand it. Thanks.

    Again, the first thing to check is does the denominator approach zero as x approaches 2. It does not. So, you can calculate the value of the limit of the rational function by just substituting in the value x = 2. The limit is 3.

  7. (From prevatth 8/30/2000) Question 8: based on the way we worked this problem in class, I thought the answer would be 5x-(x^2/500)-300. Please explain.

    In this case the limit of the denominator is zero, so you can't just substitute x = 2 in order to get the answer. The limit of the numerator is also zero. So, try and factor the numerator and cancel: One gets x^2-5x+6 = (x-2)(x-3). After cancelling, you are left with just x-3 and the limit si 2 - 3 = -1.

  8. (From bomarf 8/31/2000) Question 9:

    If x is raised to the 1/2 power, then isn't that the same as taking the square root of x? If so, then wouldn't the denominator be zero?

    Yes, on both counts. But, that doesn't mean that the value of the limit doesn't exist. Notice that the value of the numerator at x = 9 is also zero. To see what is happening, factor the numerator and cancel: x - 9 = (sqrt(x) - 3)(sqrt(x) + 3) and so when you cancel, you get sqrt(x) + 3. As x approaches 9, this approaches sqrt(9) + 3 = 6.

  9. (From prevatth 8/31/2000) Question 10:

    please explain how to find limits of x to infinity

    Calculating a limit as x approaches infinity is just a matter of looking to see what happens as x gets very large. In this case, both numerator and denominator approach infinity. To handle this case, divide numerator and denominator by x. You get (1-9/x)(2-3/x). Now look to see what happens as x gets very large. 9/x and 3/x both approach zero and so the limit is just 1/2.

  10. (From prevatth 8/31/2000) Question 12:

    please explain how to find limits of x to negative infinity and work this one out

    This is worked similarly to Question 10. When x approaches negative infinity, the numerator approaches plus infinity and the denominator approaches negative infinity. To see what is really happening, divide numerator and denominator by x^3 and arrange it as: (3x^4+x-1)/((2x-3)(6x^2-1)) = (3x +1/x^2-1/x^3)/((2-3/x)(6-1/x^2)). The limit of the denominator as x grows toward minus infinity is now clearly 2(6) = 12. The numerator approaches -infinity. So the quotient of the two approaches negative infinity.

  11. (From prevatth 8/31/2000) Question 13:

    how do you find part (d)?

    From the picture, the limit as x approaches infinity of f(x) is 1 and the limit as x approaches negative infinity of f(x) is -1. As x approaches negative infinity, -x approaches infinity and so f(-x) approaches the limit of f(x) as x approaches infinity, i.e. 1. The limit of f(x)f(-x) is the product of the two limits, which is -1 times 1, i.e. -1.

  12. (From prevatth 8/31/2000) Question 17:

    please explain this one!

    This looks nasty, but is not. First consider the quotient inside the square root sign. This is like Problem 10. To work it, divide numerator and denominator by x^5, so that it becomes (500/x^4-1/x^2+27)/(1000/x^5+3). The limit as x approaches negative infinity is 27/3 or 9. The limit of the square root is sqrt(9) = 3.

  13. (From prevatth 8/31/2000) Question 19:

    how do you do this one?

    You are trying to solve f(c) = 2, which just says x^2-10x+23=2. You can use the quadratic formula or factoring: x^2-10x+21 = (x-7)(x-3) and so x = 3 or 7 are possible values of c. Eliminate x = 7 because we are looking for a c between 2 and 6.

    For more information on the Intermediate Value Theorem, see Section 6 of the textbook.


Revised: Aug 31, 2000
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