Concept explainers
Finding Limits at Infinity In Exercises 11 and 12, find
In Exercises 11 and
(a)
(b)
(c)
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Calculus: Early Transcendental Functions
- f(x)=(1-2x)/x, find lim f(x) approaches 3 and limit f(x) approaches 4arrow_forward4. Given f(x)= { 1, if x < 0 x-1, if x ≥ 0 , estimate lim f(x) x→0- a.1 b. 0 c. does not exist d. -1 5. Given f(x)= { 1/x, if x < 4 x2, if x ≥ 4 , estimate lim f(x) x→4+ a. 1/4 b. 2 c. does not exist d. 16arrow_forwardlim x2-4x/x2-16 (Guess value of limit) x-->4arrow_forward
- f'(x) = (x-1)(x-2)(x-3)(x-4) What can be said about lim(x->infinity) if anything?arrow_forwardLet (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_forwardlim x approaches 0 (e3x-3x-1)/x2arrow_forward
- Determine value for c so that lim f(x) x--> 3 exists f(x)= 1/3x+c For x<3 -x+8 For x>3arrow_forward6. Given f(x)= { 3, if x < -2 x, if -2 ≤ x<8 x+3, if x≥8 estimate lim f(x) x→−2 a.-2 b. 1 c. does not exist d. 3 7. Using a table of values, estimate lim 1/ x-3 x→3 a. 1 b. -1 c. - ∞ d. ∞ 8. Using table of values, estimate lim x+4/ (x-5)2 x→5 a. 10 b. 8 c. - ∞ d. ∞arrow_forwarda) lim x-->negative a (f(x)) b) lim x-->postive a (f(x)) c) lim x-->a. (f(x))arrow_forward
- 1. If limit of f(x) -8/x-1=10 as x approaches 1, find limit of f(x) approaches 1 2. If limit of f(x)/x^2 = 5 , find the following limts, a) limit of f(x) as x approaches 0 and b) limit of f(x)/x as x approaches 0. 3. Show by means of an example that limit of ( f(x) +g(x) ) as x approaches zero may exist even though neither limit of f(x) approaches a nor limit of g(x) exists as x approaches aarrow_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_forwarda. lim x->1- f(x) b. lim x-> 1+ f(x)arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage