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Math Help - Limits involving a hybrid function.

  1. #1
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    I also have a bit more to ask, I apologize. These two other problems are not from my homework, but they are supplement problems from an online practice test from the same website that gives me homework and I will be getting a test on these soon and would greatly appreciate some further help. They are also limit based problems, only they are piece-wise functions. I posted my work and would like to see if I am headed in the right direction or not.



    Is this the correct way to approach this problem?

    (a) Since -6 is being approached from the left side, that would mean the value as x approaches -6 are smaller than -6, thus making me apply the first function which is g(x) = 0?

    (b) Since -6 is being approached from the right side, that means the value of x as it approaches -6 is larger than -6, so that would mean g(x) = x? And would I just leave it as x for the answer, or should I plug in the the limit -6 for x?

    (c) Since -6 is being approached ambiguously from both sides, that would make me apply the middle formula, which is g(x) = sqrt(36-x^2)? And would I plug in -6 for x, which would result with sqrt(36-(-6)^2) -> sqrt (36 + 36) = sqrt(72) = 8.485?

    (d) Now we have 6 instead of -6. If 6 is being approached from the left, that means the value of x as it approaches 6 is less than 6, which would make me apply the first function, which is g(x) = 0.

    (e) 6 is being approached from the right, which means the value of x as it approaches 6 is larger than 6. That would mean that I should apply the third function, which is g(x) = x, and we put 6 for x since that is what x is approaching.

    (f) Now we have 6 being approached from both the left and right which would imply that the middle function would be applied, which is g(x) = sqrt(36-x^2). Then sqrt(36-(6)^2) would result in sqrt(36-36) or square root of 0. Would that mean no limit exists?

    Here is another one.



    For this would I simply conclude that there is no way for me to state the value of the limit?
    Last edited by mr fantastic; September 28th 2010 at 02:56 AM. Reason: Moved new question to new thread.
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  2. #2
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    Quote Originally Posted by FullBox View Post
    I also have a bit more to ask, I apologize. These two other problems are not from my homework, but they are supplement problems from an online practice test from the same website that gives me homework and I will be getting a test on these soon and would greatly appreciate some further help. They are also limit based problems, only they are piece-wise functions. I posted my work and would like to see if I am headed in the right direction or not.



    Is this the correct way to approach this problem?

    (a) Since -6 is being approached from the left side, that would mean the value as x approaches -6 are smaller than -6, thus making me apply the first function which is g(x) = 0?

    (b) Since -6 is being approached from the right side, that means the value of x as it approaches -6 is larger than -6, so that would mean g(x) = x? And would I just leave it as x for the answer, or should I plug in the the limit -6 for x? Mr F says: You have used the wrong rule for g(x). Look again.

    (c) Since -6 is being approached ambiguously from both sides, that would make me apply the middle formula, which is g(x) = sqrt(36-x^2)? And would I plug in -6 for x, which would result with sqrt(36-(-6)^2) -> sqrt (36 + 36) = sqrt(72) = 8.485? Mr F says: 36 - (-6)^2 = 36 - 36 = 0. Be more careful with basic arithmetic. Note, however, that this calculation is relevant to (b) not (c). Note that if the limits x --> a+ and x --> a- both exist and are equal to each other, then the limit as x --> a of the function exists and is equal to that common value.

    (d) Now we have 6 instead of -6. If 6 is being approached from the left, that means the value of x as it approaches 6 is less than 6, which would make me apply the first function, which is g(x) = 0. Mr F says: The middle rule gets applied, The answer is zero.

    (e) 6 is being approached from the right, which means the value of x as it approaches 6 is larger than 6. That would mean that I should apply the third function, which is g(x) = x, and we put 6 for x since that is what x is approaching.

    (f) Now we have 6 being approached from both the left and right which would imply that the middle function would be applied, which is g(x) = sqrt(36-x^2). Then sqrt(36-(6)^2) would result in sqrt(36-36) or square root of 0. Would that mean no limit exists? Mr F says: Left hand limit and right hand limit are different therefore limit does not exist.

    Here is another one.



    For this would I simply conclude that there is no way for me to state the value of the limit?
    Left hand limit (x --> 1-) is -4. Right hand limit (x --> 1+) is a - 5. You require left hand and right hand limits to be equal if limit as x --> 1 exists ....
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  3. #3
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    FullBox, your are misunderstanding the piecewise definition of the function.
    For example, for x approaching -6 from above, you use the formula which is only correct for x> 6, not -6. If x is approaching -6 from above then we need to look at values of x that are larger than -6 but close to it- values like -5, -5.5, -5.9, etc. For those, you use the "if -6\le x\le 6" formula.

    For x approaching -6 from both sides, you do NOT just use the middle formula. You use both that and the one for x< -6. In fact, if you have already found \displaytype\lim_{x\to -6^-} f(x) and \displaytype\lim_{x\to 6^+} f(x) you don't have to do that again: if they were the same that common value is the limit, if not, there is no limit.

    For the other one, \displaytype\lim_{x\to 1^-} g(x)= \lim_{x\to 1} x^2- 5x and \displaytype\lim_{x\to 1^+} g(x)= \lim_{x\to 1} ax^3- 5. If those are the same then that is the limit. If not there is no limit. One of them will clearly depend upon "a". Perhaps there was a part of the problem that asked you to find a so that the limit does exist?
    Last edited by mr fantastic; September 28th 2010 at 02:42 PM. Reason: Replaced some itex tags with math tags.
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    Mr F says: You have used the wrong rule for g(x). Look again.
    I now understand that it should be the middle function. HallsofIvy was right, I wasn't using the definition of "limit" correctly and I was not looking at values close to -6. As you said, the answer for this one will be zero after applying the middle rule (thank you for correcting me, I didn't notice that mistake I made).

    Mr F says: 36 - (-6)^2 = 36 - 36 = 0. Be more careful with basic arithmetic. Note, however, that this calculation is relevant to (b) not (c). Note that if the limits x --> a+ and x --> a- both exist and are equal to each other, then the limit as x --> a of the function exists and is equal to that common value.
    So since both (a) and (b) are zero, that would make (c) zero as well?

    Thank you so much. I think I understand this problem now. My main issue was that when I failed to look at values near the limit and instead went all the way (for example, when 6 was approached from the left I used the formula for x <= 6, which is far too much).

    Quote Originally Posted by mr fantastic View Post
    Left hand limit (x --> 1-) is -4. Right hand limit (x --> 1+) is a - 5. You require left hand and right hand limits to be equal if limit as x --> 1 exists ....
    I'm sorry but I don't understand how you reached the conclusion for the right hand limit a to be -5.

    For the first formula, lets assume x is equal to 1.

    x^2 - 5x ----> (1)^2 - 5(1) = 1 - 5 = -4. You already stated this.

    For the second formula, lets assume x is 2 (any value greater than 1).

    a(x^3) - 5 ----> a(2^3) - 5 ---> a(8) - 5 ----> 8a-5

    If a were to be 1/8, then 8(1/8) - 5 ----> 1 - 5 = -4.

    So would that be correct for me finding a value of a that makes the limits equal?

    Quote Originally Posted by HallsofIvy View Post
    FullBox, your are misunderstanding the piecewise definition of the function.
    For example, for x approaching -6 from above, you use the formula which is only correct for x> 6, not -6. If x is approaching -6 from above then we need to look at values of x that are larger than -6 but close to it- values like -5, -5.5, -5.9, etc. For those, you use the "if -6\le x\le 6" formula.

    For x approaching -6 from both sides, you do NOT just use the middle formula. You use both that and the one for x< -6. In fact, if you have already found \displaytype\lim_{x\to -6^-} f(x) and \displaytype\lim_{x\to 6^+} f(x) you don't have to do that again: if they were the same that common value is the limit, if not, there is no limit.
    Thank you so much for pointing this out, this helped me realize my mistake. I now think I completely understand this problem.

    For the other one, \displaytype\lim_{x\to 1^-} g(x)= \lim_{x\to 1}  x^2- 5x and \displaytype\lim_{x\to 1^+} g(x)= \lim_{x\to  1} ax^3- 5. If those are the same then that is the limit. If not there is no limit. One of them will clearly depend upon "a". Perhaps there was a part of the problem that asked you to find a so that the limit does exist?
    Well it does say on the top of the problem to "determine a value for the constant a for which the limit exists", which is what I think you're asking [I'm not sure though]. As I posted above, the only thing I can think of for a to make the limits equal is 1/8, though I have a feeling I misunderstood the question once again.
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  5. #5
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    Quote Originally Posted by FullBox View Post
    I now understand that it should be the middle function. HallsofIvy was right, I wasn't using the definition of "limit" correctly and I was not looking at values close to -6. As you said, the answer for this one will be zero after applying the middle rule (thank you for correcting me, I didn't notice that mistake I made).



    So since both (a) and (b) are zero, that would make (c) zero as well? Mr F says: Yes.

    Thank you so much. I think I understand this problem now. My main issue was that when I failed to look at values near the limit and instead went all the way (for example, when 6 was approached from the left I used the formula for x <= 6, which is far too much).



    I'm sorry but I don't understand how you reached the conclusion for the right hand limit a to be -5. Mr F says: Look again. I said it was a - 5.

    For the first formula, lets assume x is equal to 1.

    x^2 - 5x ----> (1)^2 - 5(1) = 1 - 5 = -4. You already stated this.

    For the second formula, lets assume x is 2 (any value greater than 1). Mr F says: NO. You are approaching 1 not 2.

    a(x^3) - 5 ----> a(2^3) - 5 ---> a(8) - 5 ----> 8a-5

    If a were to be 1/8, then 8(1/8) - 5 ----> 1 - 5 = -4.

    So would that be correct for me finding a value of a that makes the limits equal?

    Mr F says: No. You equate the two limits x --> 1+ and x --> 1-: a - 5 = -4 => a = ....



    Thank you so much for pointing this out, this helped me realize my mistake. I now think I completely understand this problem.



    Well it does say on the top of the problem to "determine a value for the constant a for which the limit exists", which is what I think you're asking [I'm not sure though]. As I posted above, the only thing I can think of for a to make the limits equal is 1/8, though I have a feeling I misunderstood the question once again.
    ..
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