Suppose that a function f(x) is defined for all x in [-1,1]. C...

Question

Suppose that a function f(x) is defined for all x in [-1,1]. Can anything be said about the existence of limx→0 f(x)? Give reasons for your answer.

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Consider a continuous function GFT defined for all T in the interval square bracket 2.4 square bracket. Is it possible to determine the existence of the limit as T approaches 3 of GFT? Explain. So, it appears for this particular problem, we're asked to consider this continuous function of GFT, and we're trying to figure out if it's possible to determine the existence of this limit as T approaches 3 of GFT. So we're trying to explain if we can determine this existence of the specific limit using our continuous function of GFT. So with that said, now that we know what we're trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is yes, because the function is continuous within its domain, the limit as T approaches 3 exists. B is no. The existence of the limit cannot be determined without additional information about the function's behavior near T equals 3. C is yes, but only if the function is linear, and D is no because limits can only be determined for functions defined at a single point. Awesome. So with that said, first off, since GFT, so let's note that G of T is continuous, which I'll put a C in quotation marks for continuous, on the interval, square brackets 2.4, and let us quickly recall that square brackets means that it's a closed interval. If they were parenthesis, it would mean it's an open interval, but Right now we're dealing with square brackets, so that means we're dealing with a closed interval. So since GFT is continuous on this specific interval, the function does not have any breaks, jumps, or holes within this interval. So therefore, as a result, the limit of GFT as T approaches any point within this interval, including T equals 3, will exist and it will be equal to the value of the function at that point. So that means, without a doubt, the limit as T approaches 3 of G of T exists. So that means that yes, it exists. So that's a big yes. So, that means so far that's our our our part of our final answer that we're trying to solve for as we proved that this limit does indeed exist, but now we need to explain why this limit exists. So let's look through each of our multiple choice answers to see which explanation fits with the answer that we just saw for together. So let's analyze each of our multiple choice answers starting with A. So A once again is yes because the function is continuous within its domain and the limit as T approaches 3 exists, and this is correct. This this is statement, this explanation is correct because a continuous function ensures that the limit exists at every point within its domain. So without a doubt, multiple choice answer letter A. Has to be our final answer, but let's quickly explain why B, C, and D are incorrect. So, B is incorrect because continuity guarantees that the limit exists and no extra information is needed. C is incorrect because the function does not need to be linear for the limit to exist, and D is incorrect because limits describe function behavior near a point. Not just at a single point. So with that said, the only correct answer has to be multiple choice A, and once again, our final answer has to be yes, because the function is continuous within its domain. The limit as T approaches 3 exists. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Suppose that a function f(x) is defined for all x in [-1,1]. Can anything be said about the existence of limx→0 f(x)? Give reasons for your answer.

Key Concepts

Limit of a Function

Continuity

Existence of Limits

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