Concept explainers
Finding δ for a given ε using a graph Let g(x) = 2x3 − 12x2 + 26x + 4 and note that
- a. ε = 1
- b. ε = 0.5
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
Calculus: Early Transcendentals (2nd Edition)
Additional Math Textbook Solutions
Precalculus (10th Edition)
University Calculus: Early Transcendentals (3rd Edition)
Calculus & Its Applications (14th Edition)
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)
Calculus, Single Variable: Early Transcendentals (3rd Edition)
- 5. Given the graph of g(x) = {√−x−2+1, x<-2 x2+2x+5, x≥-2 below, evaluate its limit as x approaches -2. a. 3 b. 1 c. the limit does not exist d. 5 6. Evaluate lim g(x) x→-3-. a. 18 b. - ∞ c.0 d. ∞ 7. Let g(x) = -1/x2. Evaluate lim g(x) x→0 by drawing its graph. a. ∞ b. 1 c. 0 d. -∞arrow_forwarda. What is the domain of f? Express your answer in interval notation. f(x)= 1 - x^4 / x^2 - 1 b. Use a sequence of values of x near a=1 to estimate the value of limx→1 f(x). The sequence should include values such as 1.01, 1.001, etc. c. Use algebra to simplify the expression 1 - x^4 / x^2 - 1 d. True or false: f(1)=-2 e. Based on all of your work above, construct an accurate, labeled graph of y=f(x) on the interval [0,2].arrow_forwardFinding a constant Suppose Determine a value of the constant b for which lim xS2 f 1x2 exists and state the value of the limit, if possible.arrow_forward
- Proof with limit definition that: limx→1/2 (1/x)=2 I have the following: Given ε>0. choose δ=? Suppose : 0<|x-(1/2)|<δ check: |(1/x)-2| from here I do not know how to get |x-(1/2)| from |(1/x)-2| in order to find δ?arrow_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_forwardUse 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_forward
- 1. Evaluate lim (2x-5) given its graph below. x →2 a.0 b. -1 c. 2 d. 1 2. Given the graph of f(x) below, evaluate its limit as x approaches 3. a.2 b. 3 c. 1 d. 0arrow_forward(a) for g(t)= t-9/square root t-3 . Make table of values with at least six appropriate inputs to evaluate the limits as t tends to 9. (b) make a conjecture about the value of lim t-9/square root t-3. t->9arrow_forwardlim x->0 ((x+k)2 -x2/k)arrow_forward
- 3. The graph of f(x)= { 3, x≤-1 (x-1)2, x>-1 is shown below. Evaluate lim f(x) x→-1-. a. The limit does not exist b. -1 c. 3 d. 4 4. Given the graph of f(x) below, evaluate its limit as x approaches 2 from the right. a- ∞ b. The limit does not exist c. 2 d. 4arrow_forwardA positive number _ and the limit L of a function f at a are given. Find a number δ such that|f(x) − L| < _ if 0 < |x − a| < δ. limx→3 x^2−9/x−3= 6; ϵ = 0.05arrow_forwardThe greatest integer function, also known as the floor function, isdefined by [ x ] = n, where n is the unique integer such that n < x < n + 1.Sketch the graph of y = [ x ] . Calculate for c an integer:(a) lim [ x ] (b) lim [ x ] ( c) lim [ x ] x→ c x→ c+ x→ 2.6arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning