Problem 18 - Thomas Calculus

Question

Find the limits in Exercises \(11-18\) $$a. \lim _{x \rightarrow 1^{+}} \frac{\sqrt{2 x}(x-1)}{|x-1|} \quad b. \lim _{x \rightarrow 1} \frac{\sqrt{2 x}(x-1)}{|x-1|}$$

Step-by-Step Solution

Answer

a) \( \sqrt{2} \); b) Does not exist.

Step 1: Analyze the expression for limit (a)

For the limit \( \lim_{x \rightarrow 1^{+}} \frac{\sqrt{2x}(x-1)}{|x-1|} \), realize that as \( x \) approaches 1 from the right, \( x-1 \) is positive. Therefore, \(|x-1| = x-1\). This simplifies the expression to \( \frac{\sqrt{2x}(x-1)}{x-1} \).

Step 2: Simplify the expression for limit (a)

By simplifying \( \frac{\sqrt{2x}(x-1)}{x-1} = \sqrt{2x} \) because \( (x-1) \) in the numerator and denominator cancels out. As \( x \rightarrow 1^{+} \), \( \sqrt{2x} \rightarrow \sqrt{2} \cdot 1 = \sqrt{2} \).

Step 3: Compute the limit (a)

The limit \( \lim_{x \rightarrow 1^{+}} \sqrt{2x} \) simplifies as \( x \rightarrow 1 \). Thus, the limit is \( \sqrt{2} \).

Step 4: Analyze the expression for limit (b)

For \( \lim_{x \rightarrow 1} \frac{\sqrt{2x}(x-1)}{|x-1|} \), consider both the right and left limits since it involves an approaching point. When \( x \rightarrow 1^{+} \), we already know the simplification from previous steps is \( \sqrt{2} \).

Step 5: Consider left-hand limit for (b)

For \( x \rightarrow 1^{-} \), \( x-1 \) is negative, thus \(|x-1| = -(x-1)\). Substituting this gives \( \frac{\sqrt{2x}(x-1)}{-(x-1)} = -\sqrt{2x} \).

Step 6: Evaluate the left-hand limit for (b)

As \( x \rightarrow 1^{-} \), \( -\sqrt{2x} \rightarrow -\sqrt{2} \cdot 1 = -\sqrt{2} \).

Step 7: Compute the bidirectional limit (b)

The right-hand and left-hand limits are \( \sqrt{2} \) and \( -\sqrt{2} \) respectively. These are not equal, indicating that \( \lim_{x \rightarrow 1} \frac{\sqrt{2x}(x-1)}{|x-1|} \) does not exist.

Key Concepts

One-sided limits

One-sided limits are a useful concept when exploring how a function behaves as it approaches a particular point from one direction. In the provided exercise, we examine the limit as \( x \rightarrow 1^{+} \) and \( x \rightarrow 1^{-} \).

• \( x \rightarrow 1^{+} \) means that \( x \) is approaching 1 from the right side, using very slightly larger values than 1.
• \( x \rightarrow 1^{-} \) indicates \( x \) is approaching 1 from the left, with very slightly smaller values than 1.

To understand these limits, note how the expression changes based on direction. In the exercise:
* At \( x \rightarrow 1^{+} \), \( |x-1| = x-1 \) because values are positive, simplifying \( \frac{\sqrt{2x}(x-1)}{|x-1|} \) to \( \sqrt{2x} \).* At \( x \rightarrow 1^{-} \), \( |x-1| = -(x-1) \), making it different from the right direction and yielding \( -\sqrt{2x} \).
Understanding how limits change directionally helps in characterizing function continuity and analyzing points where limits might not exist.

Limit expression analysis

Analyzing a limit involves breaking down the expression to determine its behavior as \( x \) approaches a specific value. For the exercise, the original expression is \( \frac{\sqrt{2x}(x-1)}{|x-1|} \).
Initially, it might appear complex, but here are the steps to analyze it simpler:

• Identify the form of \( |x-1| \): realize the absolute value expression changes depending on whether \( x \) approaches from the right or left.
• Simplify when possible: once \( |x-1| \) is expressed correctly for a direction, cancel like terms.
• Determine the resultant expression behavior as \( x \) precisely approaches the given point, making it straightforward like \( \sqrt{2x} \) or \( -\sqrt{2x} \).

This particular exercise illustrates how expression analysis simplifies a seemingly complex function into an easy-to-assess limit using elementary arithmetic and logical judgements.

Indeterminate forms

Indeterminate forms often occur when evaluating limits and can be a sign that the limit is not straightforward. Common forms include \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where the result is not immediately clear.
In this exercise, the issue arises when trying to evaluate \( \lim_{x \rightarrow 1} \frac{\sqrt{2x}(x-1)}{|x-1|} \). As \( x \rightarrow 1 \), both the numerator and denominator tend towards zero:

• The numerator \( \sqrt{2x}(x-1) \) becomes 0 because \( x-1 \rightarrow 0 \).
• The denominator \( |x-1| \) also tends towards 0, depending on the direction \( x \) approaches 1.

Since the limits from the right (\( \sqrt{2} \)) and left (-\( \sqrt{2} \)) are not the same, the indeterminate form highlights that the limit does not exist. Recognizing indeterminate forms allows students to know when to further analyze or use special techniques like L'Hospital's Rule, algebraic manipulation, or examining one-sided limits to resolve them.