# Thread: Limits involving a hybrid function.

1. 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?

2. Originally Posted by FullBox
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 ....

3. 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 $\displaystyle -6\le x\le 6$" 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 $\displaystyle \displaytype\lim_{x\to -6^-} f(x)$ and $\displaystyle \displaytype\lim_{x\to 6^+} f(x)$ 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, $\displaystyle \displaytype\lim_{x\to 1^-} g(x)= \lim_{x\to 1} x^2- 5x$ and $\displaystyle \displaytype\lim_{x\to 1^+} g(x)= \lim_{x\to 1} ax^3- 5$. 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?

4. 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).

Originally Posted by mr fantastic
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?

Originally Posted by HallsofIvy
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 $\displaystyle -6\le x\le 6$" 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 $\displaystyle \displaytype\lim_{x\to -6^-} f(x)$ and $\displaystyle \displaytype\lim_{x\to 6^+} f(x)$ 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, $\displaystyle \displaytype\lim_{x\to 1^-} g(x)= \lim_{x\to 1} x^2- 5x$ and $\displaystyle \displaytype\lim_{x\to 1^+} g(x)= \lim_{x\to 1} ax^3- 5$. 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.

5. Originally Posted by FullBox
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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