Math Problem Statement

Analyze the graph of the function f(x) to evaluate one-sided limits and determine whether the statement lim(x → 0) f(x) = 2 is true or false.

Solution

From the image, we are tasked with analyzing the graph of the function f(x)f(x) to evaluate some one-sided limits and verify a statement.

1. The First Limit:

The limit as xx approaches 00^- (from the left) is already provided: limx0f(x)=2\lim_{{x \to 0^-}} f(x) = 2 This means as xx approaches 0 from the left, the function value approaches 2, which matches the graph's behavior near x=0x = 0.

2. The Second Limit:

We now need to determine: limx0+f(x)\lim_{{x \to 0^+}} f(x) Looking at the graph, as xx approaches 0 from the right, the function value still seems to approach 22 (there’s no break or jump near x=0x = 0 on the right side). Therefore: limx0+f(x)=2\lim_{{x \to 0^+}} f(x) = 2

3. Statement Verification:

The next task is to determine whether the statement limx0f(x)=2\lim_{{x \to 0}} f(x) = 2 is True or False. Since both the left-hand limit (x0x \to 0^-) and the right-hand limit (x0+x \to 0^+) are equal to 2, the overall limit exists and is equal to 2. Thus, the statement is True.


Would you like further details or clarification?

Here are five related questions to expand your understanding:

  1. What happens to f(x)f(x) when x4x \to 4^- and x4+x \to 4^+?
  2. How can you determine if a function is continuous at a point using limits?
  3. What is the relationship between one-sided limits and the overall limit?
  4. How can you identify discontinuities in a graph based on limits?
  5. What is the difference between limits that exist and limits that do not?

Tip: When analyzing limits graphically, always check the function's behavior from both sides of the point in question to ensure the left-hand and right-hand limits match.

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Math Problem Analysis

Mathematical Concepts

Limits
One-sided limits
Continuity

Formulas

lim(x → 0⁻) f(x) = 2
lim(x → 0⁺) f(x) = 2
lim(x → 0) f(x) = 2

Theorems

Limit Definition
Continuity at a Point

Suitable Grade Level

Grades 11-12