Math Problem Statement

Question content area left Part 1 Which of the statements a through k about the function yequals​f(x) graphed here are​ true, and which are​ false? -4 -2 2 4 -4 -2 2 4 x y

A coordinate system has a horizontal x-axis that ranges from negative 4 to 4 and a vertical y-axis that ranges from negative 4 to 4. A graph is formed from a series of line segments. From left to right, a line segment begins at the closed circle (~xpt1, ~ypt1), moves to the open circle at (~xpt2, ~ypt2), and ends at the closed circle (~xpt3, ~ypt3). A point is plotted at (0, 1). Another line segment begins at the closed circle (2, 0) and ends at the closed circle (3, 1). . . . Question content area right Part 1 a. The statement ModifyingBelow lim With x right arrow 0 f left parenthesis x right parenthesis exists is true. Part 2 b. The statement ModifyingBelow lim With x right arrow 0 f left parenthesis x right parenthesisequals0 is true. Part 3 c. The statement ModifyingBelow lim With x right arrow 0 f left parenthesis x right parenthesisequals1 is ▼ true. false.

Solution

Based on the description of the graph, I'll go through each statement one by one and determine whether it is true or false.

Statement a:

"The statement limx0f(x)\lim_{{x \to 0}} f(x) exists is true."

  • This statement refers to whether the limit of f(x)f(x) as xx approaches 0 exists. A limit exists if the function approaches the same value from both the left and the right sides as x0x \to 0.
  • From the description, we know that at x=0x = 0, there is a point at (0,1)(0, 1), but the behavior of the graph near x=0x = 0 is not fully specified. If the graph approaches the same value from both sides, the limit exists. Since there's no information about a discontinuity at x=0x = 0, we assume the limit exists.

Conclusion: True (the limit exists, based on the provided description).


Statement b:

"The statement limx0f(x)=0\lim_{{x \to 0}} f(x) = 0 is true."

  • The limit as x0x \to 0 would equal 0 if the function approaches the value 0 from both sides of x=0x = 0. However, it is given that a point is plotted at (0,1)(0, 1), which means the function value at x=0x = 0 is 1, not 0.
  • Since the function is continuous at x=0x = 0 (i.e., the value at (0,1)(0, 1) is the limit from both sides), the limit does not equal 0.

Conclusion: False (the limit is not 0).


Statement c:

"The statement limx0f(x)=1\lim_{{x \to 0}} f(x) = 1 is true."

  • Given that the graph passes through the point (0,1)(0, 1), we can infer that the function value at x=0x = 0 is 1.
  • If the function approaches the value of 1 from both sides as xx approaches 0, then the limit as x0x \to 0 would indeed be 1.

Conclusion: True (the limit equals 1, based on the provided description).


Summary of answers:

  • a. True: limx0f(x)\lim_{{x \to 0}} f(x) exists.
  • b. False: limx0f(x)=0\lim_{{x \to 0}} f(x) = 0.
  • c. True: limx0f(x)=1\lim_{{x \to 0}} f(x) = 1.

Would you like to dive deeper into any of these statements, or do you have further questions? Here are some related questions you might find interesting:

  1. What does it mean for a limit to "exist" at a specific point?
  2. How do we determine if a function is continuous at a point?
  3. How does an open or closed circle on a graph affect the limit of a function?
  4. What are some examples of limits that do not exist?
  5. What is the significance of the behavior of a function near discontinuities?

Tip: When evaluating limits, always check both the left-hand limit (limxc\lim_{{x \to c^-}}) and right-hand limit (limxc+\lim_{{x \to c^+}}) separately. The limit exists only if both one-sided limits agree.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Piecewise Functions

Formulas

lim (x → c) f(x)

Theorems

Limit Existence
Continuity Theorem

Suitable Grade Level

Grades 10-12