Explain what it means to say that lim x → 9− f(x) = 3 and lim x → 9+ f(x) = 5. As x approaches 9, f(x) approaches 5, but f(9) = 3. As x approaches 9 from the right, f(x) approaches 3. As x approaches 9 from the left, f(x) approaches 5. As x approaches 9, f(x) approaches 3, but f(9) = 5. As x approaches 9 from the left, f(x) approaches 3. As x approaches 9 from the right, f(x) approaches 5. In this situation is it possible that lim x → 9 f(x) exists? Explain. Yes, f(x) could have a hole at (9, 3) and be defined such that f(9) = 5. Yes, f(x) could have a hole at (9, 5) and be defined such that f(9) = 3. Yes, if f(x) has a vertical asymptote at x = 9, it can be defined such that lim x→9− f(x) = 3, lim x→9+ f(x) = 5, and lim x→9 f(x) exists. No, lim x→9 f(x) cannot exist if lim x→9− f(x) ≠ lim x→9+ f(x).