Concept explainers
Finding Limits at Infinity In Exercises 11 and 12, find
(a)
(b)
(c)
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Calculus of a Single Variable
- a = 0 x −0.1 −0.01 −0.001 −0.0001 0 0.0001 0.001 0.01 0.1 f(x) 19.0303 −3.00303 −3.0003 −3.00003 0 3.00003 3.0003 3.00303 3.0303 (a) lim x→a− f(x) (b) lim x→a+ f(x)arrow_forward(Right and Left Limits). Introductory calculus coursestypically refer to the right-hand limit of a function as the limit obtained by“letting x approach a from the right-hand side.” (a) Give a proper definition in the style of Definition 4.2.1 ((Functional Limit).for the right-hand and left-hand limit statements: limx→a+f(x) = L and limx→a−f(x) = M. (b) Prove that limx→a f(x) = L if and only if both the right and left-handlimits equal L.arrow_forwardlim x approaches -3 x + 3 /[x2 + 4x + 3]arrow_forward
- Let (x) = |x - 2| / x - 2 A) what is the domain of g(x)? B) Use numerical methods to find lim x—> 2- g(x) and lim x—> 2+ g(x). C) based on your answer to (b), what is lim x—> 2 g(x)? D) sketch an accurate graph of g(x) on the interval [-4,4]. Be sure to include any needed open or closed circles.arrow_forwardconsidering the function: I. The value of a so that the limit of f(x) exists when x tends to 0 is: a= 2. the function( ) and keeps going( ) has a jump-type discontinuity( ) has a removable type discontinuity( ) has an infinite type discontinuityarrow_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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