Math Problem Statement

Evaluate the limit: \( \lim_{{x \to 2}} \frac{|x - 2|}{x - 2} \)

Solution

The given mathematical expression is:

limx2x2x2\lim_{{x \to 2}} \frac{|x - 2|}{x - 2}

This is a piecewise function, so to evaluate the limit, we need to consider the behavior of the expression as xx approaches 2 from both the left and the right.

Case 1: x2x \to 2^- (Approaching from the left)

When x<2x < 2, the absolute value function x2|x - 2| becomes (x2)-(x - 2). So the expression becomes:

x2x2=(x2)x2=1\frac{|x - 2|}{x - 2} = \frac{-(x - 2)}{x - 2} = -1

Thus, as xx approaches 2 from the left, the value of the function is 1-1.

Case 2: x2+x \to 2^+ (Approaching from the right)

When x>2x > 2, the absolute value function x2|x - 2| simplifies to x2x - 2. So the expression becomes:

x2x2=(x2)x2=1\frac{|x - 2|}{x - 2} = \frac{(x - 2)}{x - 2} = 1

Thus, as xx approaches 2 from the right, the value of the function is 11.

Conclusion:

Since the left-hand limit is 1-1 and the right-hand limit is 11, the two limits do not agree. Therefore, the limit does not exist:

limx2x2x2does not exist.\lim_{{x \to 2}} \frac{|x - 2|}{x - 2} \quad \text{does not exist}.

Would you like further details or explanations? Here are 5 related questions to explore:

  1. What is a one-sided limit and how does it differ from a two-sided limit?
  2. How does the absolute value function affect limits?
  3. Can limits exist at points of discontinuity?
  4. How would this limit change if the expression were slightly modified, e.g., limx3x3x3\lim_{{x \to 3}} \frac{|x - 3|}{x - 3}?
  5. What is the relationship between continuity and the existence of limits?

Tip: When dealing with limits involving absolute values, always check the behavior from both sides of the point in question.

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Math Problem Analysis

Mathematical Concepts

Limits
Piecewise Functions
Absolute Value

Formulas

\( \lim_{{x \to c}} f(x) \)
\( |x - a| \)

Theorems

Limit Theorems
Left-Hand and Right-Hand Limits

Suitable Grade Level

Grades 11-12