For the graph shown, where does the limit of f(x) not exist?4+3+2-5 4 3-2-44 2 3 4 5-2-3-4-5limf(x)lim f(x)o limf(x)limf (x)6 7 89.

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Answered: For the graph shown, where does the…

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter5: A Survey Of Other Common Functions

Section5.6: Higher-degree Polynomials And Rational Functions

Problem 3TU

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8.Given the graph of f(x) below, evaluate its limit as x approaches 2 from the left. a. 2 b. - ∞ c. 0 d. -2 9. Evaluate lim g(x).                     x →-3- a. 18 b. - ∞    c.0    d. ∞
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
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. 0
Evaluate the limit numerically. lim?→x+7/x^2+x-42 = ?? x -> -7 f(-7.1) = ? f(-7.01) = ? f(-7.001) = ? f(-6.9) = ? f(-6.99) = ? f(-6.999) = ?
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Let f(x) = x² + 4 and g(x) = 3x. Find the limit limit lim g(f(x)). and limit →5. The anser is C, but I am not getting that answer.
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Use the graph to find the limit L (if it exists). If the limit does not exist, explain why. (If an answer does not exist, enter DNE.) g(x) = −2x x − 2 (a) lim x→2 g(x) L = The limit does not exist at x = 2 because the function does not approach f(2) as x approaches 2.The limit does not exist at x = 2 because the function is not continuous at any x value. The limit does not exist at x = 2 because f(2) ≠ 2.The limit does not exist at x = 2 because the function increases and decreases without bound as x approaches 2.The limit exists at x = 2. (b) lim x→0 g(x) L = The limit does not exist at x = 0 because the function does not approach f(0) as x approaches 0.The limit does not exist at x = 0 because the function is not continuous at any x value. The limit does not exist at x = 0 because f(0) ≠ 0.The limit does not exist at x = 0 because the function increases and decreases without bound as x approaches 0.The limit exists at x = 0.
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