Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

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) \(\lim _{x \rightarrow 1^{-}}\left[\frac{1}{x-1}-\frac{1}{|x-1|}\right]\)

Step-by-Step Solution

Answer

(a) Does not exist. (b) -1. (c) -3. (d) \(-\infty\).

Step 1: Analyze the First Sub-Problem

To solve \( \lim _{x \rightarrow 1} \frac{|x-1|}{x-1} \), we must consider the behavior of the function as \( x \) approaches 1 from both the left (\( x \rightarrow 1^- \)) and the right (\( x \rightarrow 1^+ \)). If \( x \) approaches 1 from the left, then \( x-1 \) is negative, so \( |x-1| = -(x-1) \). On the other hand, as \( x \) approaches from the right, \( x-1 \) is positive, hence \( |x-1| = x-1 \).

Step 2: Evaluate from the Left for (a)

Evaluate \( \lim _{x \rightarrow 1^-} \frac{|x-1|}{x-1} \). For \( x < 1 \), the expression \( |x-1| = -(x-1) \). Thus, the function becomes \( \frac{-(x-1)}{x-1} = -1 \).

Step 3: Evaluate from the Right for (a)

Evaluate \( \lim _{x \rightarrow 1^+} \frac{|x-1|}{x-1} \). For \( x > 1 \), \( |x-1| = x-1 \). So, the function becomes \( \frac{x-1}{x-1} = 1 \).

Step 4: Conclude for (a)

The left-hand limit is \(-1\) and the right-hand limit is \(1\). Since the limits from the left and right do not match, \( \lim _{x \rightarrow 1} \frac{|x-1|}{x-1} \) does not exist.

Step 5: Evaluate the Second Sub-Problem

For \( \lim _{x \rightarrow 1^{-}} \frac{|x-1|}{x-1} \), since this is a one-sided limit as \( x \to 1^- \), \( x-1 < 0 \) implies \( |x-1| = -(x-1) \). Thus, the expression simplifies to \( \frac{-(x-1)}{x-1} = -1 \).

Step 6: Analyze Sub-Problem (c)

Consider \( \lim _{x \rightarrow 1^{-}} \frac{x^{2} - |x-1| - 1}{|x-1|} \). As \( x \to 1^- \), \( |x-1| = -(x-1) \). Substitute this into the expression to get \( \frac{x^2 + (x-1) - 1}{-(x-1)} \).

Step 7: Simplify Expression for (c)

The expression simplifies to \( \frac{x^2 + x - 2}{-(x-1)} \). First, factor the numerator: \( (x-1)(x+2) \), so it becomes \( -\frac{(x-1)(x+2)}{x-1} \). Cancel \( x-1 \) (since \( x eq 1 \)) giving \( -(x+2) \). Evaluate at \( x = 1^- \): \( -(1+2) = -3 \).

Step 8: Analyze Sub-Problem (d)

For \( \lim _{x \rightarrow 1^{-}} \left[ \frac{1}{x-1} - \frac{1}{|x-1|} \right] \), we note that \( |x-1| = -(x-1) \) as \( x \rightarrow 1^- \). Hence, the expression becomes \( \frac{1}{x-1} - \frac{1}{-(x-1)} = \frac{1}{x-1} + \frac{1}{x-1} \).

Step 9: Evaluate Simplified Expression for (d)

So the expression simplifies to \( \frac{2}{x-1} \). Since \( x - 1 \to 0^- \), \( \frac{2}{x-1} \to -\infty \) as \( x \to 1^- \).

Key Concepts

One-Sided Limits

When working with limits, sometimes we only want to know the behavior of a function as it approaches a specific point from one side, either from the left or the right.
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 \( |x-1| = x-1 \).

These subtle differences are crucial to solving problems where behavior changes based on the direction of approach.

Discontinuous Functions

Functions are not always smooth and continuous at every point. Wherever there are breaks, jumps, or holes in the graph, a function is said to be discontinuous.
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 \( (x \to 1^{+}) \).

The differing limits indicate a jump discontinuity at \( x=1 \), rendering the overall limit nonexistent at this point.

Absolute Value Properties

Absolute value essentially measures the distance of a number from zero, always resulting in a non-negative number.
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 \( |x-1| = x-1 \).

This property makes absolute values instrumental in evaluating one-sided limits, as shown in the examples above, where we need to carefully substitute \(|x-1|\) based on the sign of the expression inside the absolute value.

Factoring and Simplifying Expressions

Algebraic manipulation is often necessary to simplify expressions and find limits effectively.
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 \( -(x+2) \).

By plugging \( x \to 1^{-} \) into this simplified form, we easily find that the limit is \(-3\).
Such techniques help handle complex expressions and make solving calculus problems more manageable.