Suppose g(x)=f(1−x) for all x, lim x→1^+ f(x)=4, and lim x→1^−...

Question

Suppose g(x)=f(1−x) for all x, lim x→1^+ f(x)=4, and lim x→1^− f(x)=6. Find lim x→0^+ g(x) and lim x→0^− g(x).

Accepted Answer

Welcome back, everyone, to another video. Consider H of X, which is equal to K 3 minus X, for all X, where the limit of KXX approaches 2 from the right is equal to 5, and the limit of KX as X approaches 2 from the left is equal to 7, determine the limit of HX as X approaches 1 from the right and from the left. And we're given 4 answer choices. A says 7 and 5 respectively. B 5 and 7 respectively, C 12 and 7 respectively, and D 5 and 12 respectively. So first of all, what we're going to do is simply identify the limit that we want to evaluate. We want to evaluate the limit as X approaches one from the right of H of X. And now if we use the definition, well, we are evaluating the limit as x approaches one from the right of k of 3 minus X. This expression is not, is not really useful because, well, we're given two limits of K of X and in this case we're given k of 3 minus X. So what we're going to do is simply manipulate that and we're going to say that there is some variables C which is equals to 3 minus X. And why are we really doing this? Well, we're going to get a limit of K of Z, right? Because 3 minus X is Z. So we're basically changing the limits themselves, right? We're going to look at where Z goes. So if Z is equals to 3 minus X and X approaches 1 from the right, well, what can we tell? Essentially what we're going to do is simply say that if X is equal to 1, then Z is going to be equal to 23 minus 1, right? But C is going to approach 2 from which side. Well, if we are approaching X from the right, this means that. X is greater than one because we're approaching one from the right. And if 1 is, if we are approaching 1 from the right, X is greater than 1, subtracting a value that is greater than 1 from 3 gives us a value that is less than 2. And a value that is less than 2 can be described by Z approaches 2 from the left. So in this case, we get a limit of K of Z as Z approaches 2 from the left. And now this is consistent with one of our limits, which is equal to 7. We're just using a different variable in this case Z instead of X, but this doesn't change the fact that the limit is going to be equal to 7. And we can clearly see that the correct answer choice is A, right, because the first one is 7, but now we're just going to prove that the other limit is 5. So we get limit as X approaches 1 from the left of each of X. repeating all of those steps, we are going to get the limit as X approaches 1 from the left of K of 3 minus X. And then we're just going to use a different variable. Let's suppose that we use W. So we get the limit of K of W, and now we want to see where W goes. So, in this case, W simply defines 3 minus X. Once again, we know that X in this case is equal to 1, then W is equal to 2. And from here we can say that W tends to 2 from the right. Why is this the case? Well, X is less than 1 because we're approaching 1 from the left. And if we're subtracting a value that is less than 1 from 3, we get a value that is greater than 2. So we are approaching 2 from the right. So we can say that W approaches 2 from the right. And this is consistent with the other limit which is equal to 5 in the given problem. And that said, we can conclude that the correct answer to this problem corresponds to the answer choice A. Thank you for watching.

Suppose g(x)=f(1−x) for all x, lim x→1^+ f(x)=4, and lim x→1^− f(x)=6. Find lim x→0^+ g(x) and lim x→0^− g(x).

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