Concept explainers
(a) What is wrong with the following equation?
(b) In view of part (a), explain why the equation
is correct
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Calculus (MindTap Course List)
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- 3. Evaluate lim x→5 √x+4-3/ x-5. Show your complete solution. 4. Evaluate lim x→1 x-1/ √x-1. Show your complete solution. 5. Prove that lim t→-2 √1-t3/t +3/2 / t+2= 1/4. Show your complete solution.arrow_forwardlim x approaches 1 .... [(x^-1) - 1]/(x -1)arrow_forwardGiven lim (4x-3) as x approaches 1. Use the definition of a limt to find a number delta such that the absolute value of x-a is less than delta when the absolute value of f(x)-L is less than 0.08arrow_forward
- How to evaluate this? lim 5x + 2 x→-1 lim 3x²-6x+1 x→4 lim √x²-5x+7 x→2 lim (-6x+1/8)³ x→1 lim x³+2x²-3x²-2 / x³-2x+1 x→0arrow_forwarda) value of f(1) b) lim x→1-f(x) c) lim x→1+f(x) d) Does lim x→1 f(x) exist? If so, find value. If not, explain why. e) lim x→2+f(x) f) lim x→2-f(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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