Chapter 2: Problem 43
Find each of the following limits or state that it does not exist. (a) \(\lim _{x \rightarrow 1} \frac{|x-1|}{x-1}\) (b) \(\lim _{x \rightarrow 1^{-}} \frac{|x-1|}{x-1}\) (c) \(\lim _{x \rightarrow 1^{-}} \frac{x^{2}-|x-1|-1}{|x-1|}\) (d) \(\lim _{x \rightarrow 1^{-}}\left[\frac{1}{x-1}-\frac{1}{|x-1|}\right]\)
Short Answer
Step by step solution
Analyze the First Sub-Problem
Evaluate from the Left for (a)
Evaluate from the Right for (a)
Conclude for (a)
Evaluate the Second Sub-Problem
Analyze Sub-Problem (c)
Simplify Expression for (c)
Analyze Sub-Problem (d)
Evaluate Simplified Expression for (d)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
One-Sided Limits
These are called one-sided limits. A one-sided limit as you approach from the left is denoted with a negative superscript, for example, \( \lim_{x \to 1^{-}} \), and from the right with a positive superscript, \( \lim_{x \to 1^{+}} \).
With one-sided limits, we're focusing on just a small slice of the graph near the point. In problems like \( \lim_{x \to 1^{-}} \frac{|x-1|}{x-1} \), we're interested only in values where \( x \) is slightly less than 1.
- For \( x \to 1^{-} \): Consider when \( x \) approaches 1 from values smaller than 1. Thus, \( x-1 \) is negative, so \( |x-1| = -(x-1) \).
- For \( x \to 1^{+} \): Consider when \( x \) approaches 1 from values larger than 1. Here, \( x-1 \) is positive, making \( |x-1| = x-1 \).
Discontinuous Functions
One key feature of discontinuous functions is that different limits from left and right at a certain point may not match.
For instance, the function \( \frac{|x-1|}{x-1} \) is discontinuous at \( x = 1 \) because:
- The left-hand limit is \(-1\) when approaching from the left \( (x \to 1^{-}) \).
- The right-hand limit is \(1\) when approaching from the right \( (x \to 1^{+}) \).
Absolute Value Properties
It plays a crucial role in calculus, especially when considering the direction of approach in limits.
For the expression \( |x-1| \), its value changes depending on whether \( x \) is greater than or less than 1.
- If \( x < 1 \), then \( x-1 \) is negative, and thus \( |x-1| = -(x-1) \).
- If \( x > 1 \), \( x-1 \) is positive, so \( |x-1| = x-1 \).
Factoring and Simplifying Expressions
One common technique is factoring, where we break down a complex expression into simpler parts that are easier to work with.
Consider \( \lim_{x \to 1^{-}} \frac{x^2 - |x-1| - 1}{|x-1|} \).
- First, recognize \( |x-1| = -(x-1) \) for \( x \to 1^{-} \). Substitute this into the expression: \( \frac{x^2 + (x-1) - 1}{-(x-1)} \).
- Simplify by factoring the numerator: \( x^2 + x - 2 = (x-1)(x+2) \).
- This allows us to cancel \( x-1 \) in the denominator and numerator (since \( x eq 1 \)) leading to the expression \( -(x+2) \).
Such techniques help handle complex expressions and make solving calculus problems more manageable.