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One-sided and two-sided limits Let
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Single Variable Calculus: Early Transcendentals (2nd Edition) - Standalone book
- Use properties of limits and algebraic methods to find the limit, if it exists. lim x→−3 x2 − 9 x + 3 Step 1 We want to use properties of limits and algebraic methods to find lim x→−3 x2 − 9 x + 3 . Note that the function is a function. The numerator and denominator are 0 at x = , and thus we have the indeterminate form at x = . We can factor from the numerator and reduce the fraction. lim x→−3 x2 − 9 x + 3 = lim x→−3 (x − 3) x + 3 = lim x→−3 = − 3 =arrow_forwardlim x2-4x/x2-16 (Guess value of limit) x-->4arrow_forwardA. Find lim x->4^+ and lim x->4^- B. Does lim x->4 f(x) exist C. Find lim x->5^+ & lim x->5^- D. Does lim x->5 f(x) existarrow_forward
- a) 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_forwarddetermine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If lim x→7 g(x) = 0 and lim x→7 f(x)/g(x) = 0 exists, then lim x→7 f(x) = 0. True. If lim x→7 f(x) is not equal to zero and lim x→7 g(x) = 0, then lim x→7 f(x)/g(x) does not exist. True. Any number divided by zero is equal to zero. False. Let g(x) = (x − 7) and f(x) = (x − 1)(x − 7). Then lim x→7 g(x) = 0 and lim x→7 f(x)/g(x) = 0 exists, but lim x→7 f(x) is not equal to 0. False. Divison by zero is not allowed. False. There is not enough information given to determine lim x→7 f(x)/g(x)arrow_forwardA wrong statement about limits Show by example that thefollowing statement is wrong.The number L is the limit of ƒ(x) as x approaches cif ƒ(x) gets closer to L as x approaches c.Explain why the function in your example does not have thegiven value of L as a limit as xS c.arrow_forward
- using the definition of continuity and the properties of limits to show that the function is continuous at the given number a. f(x)= 3x2+(x+2)5, a= -1 can you please explain this problem with workarrow_forwardlim x1/(1-x) x -> 1+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
- Functions and Change: A Modeling Approach to Coll...AlgebraISBN:9781337111348Author:Bruce Crauder, Benny Evans, Alan NoellPublisher:Cengage LearningCollege AlgebraAlgebraISBN:9781305115545Author:James Stewart, Lothar Redlin, Saleem WatsonPublisher:Cengage LearningAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage