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 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?