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Calculus: Early Transcendentals (3rd Edition)
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- lim. 3cosx x—> infinityarrow_forwardLim x approaches to - infinity (5-1/x)arrow_forwardlim x → 9− f(x) = 2 and lim x → 9+ f(x) = 4. As x approaches 9 from the right, f(x) approaches 2. As x approaches 9 from the left, f(x) approaches 4. As x approaches 9 from the left, f(x) approaches 2. As x approaches 9 from the right, f(x) approaches 4. As x approaches 9, f(x) approaches 4, but f(9) = 2. As x approaches 9, f(x) approaches 2, but f(9) = 4. In this situation is it possible that lim x → 9 f(x) exists? Explain. Yes, f(x) could have a hole at (9, 2) and be defined such that f(9) = 4. Yes, f(x) could have a hole at (9, 4) and be defined such that f(9) = 2. Yes, if f(x) has a vertical asymptote at x = 9, it can be defined such that lim x→9− f(x) = 2, lim x→9+ f(x) = 4, 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).arrow_forward
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