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Explain what it means to say that lim ((x) = 3 and lim A(x) = 5. x - 2 x - 2 As x approaches 2, f(x) approaches 5, but /(2) = 3. ( As x approaches 2 from the right, A(x) approaches 3. As x approaches 2 from the left, f(x) approaches 5. As x approaches 2 from the left, A(x) approaches 3. As x approaches 2 from the right, A(x) approaches 5. As x approaches 2, f(x) approaches 3, but /(2) = 5. In this situation is it possible that lim ((x) exists? Explain. x - Yes, A(x) could have a hole at (2, 3) and be defined such that /(2) = 5. O Yes, A(x) could have a hole at (2, 5) and be defined such that /(2) = 3. Yes, if f(x) has a vertical asymptote at x = 2, it can be defined such that lim ((x) = 3, lim f(x) = 5, and lim A(x) exists. ( No, lim f(x) cannot exist if lim ((x) # lim A(x). x-2 x -2 x-2