Problem 68 - Calculus: Early Transcendentals

Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 5} \frac{|x-5|}{x^{2}-25}$$

Step-by-Step Solution

Answer

Answer: The limit of the function as x approaches 5 is $$\frac{1}{10}$$.

Step 1: Identify the Limit Definition

To find the limit of the function as x approaches 5, we want to see what happens to the function when x gets very close to 5 but not equal to 5. $$\lim_{x \rightarrow 5} \frac{|x-5|}{x^{2}-25}$$

Step 2: Absolute Value Properties

Recall that the absolute value of a number has two cases: 1. If the number inside the absolute value is positive, then the absolute value is the number itself. 2. If the number inside the absolute value is negative, then the absolute value is the opposite of the number. Therefore, |x-5| has two cases: 1. If \(x > 5\), then \(|x-5| = x - 5\). 2. If \(x < 5\), then \(|x-5| = -(x - 5)\). We will analyze the limit for each case separately.

Step 3: Case 1 - x > 5

For the case where \(x > 5\), we have: $$\lim_{x \rightarrow 5^{+}} \frac{x-5}{x^{2}-25}$$ Factor the denominator to get: $$\lim_{x \rightarrow 5^{+}} \frac{x-5}{(x-5)(x+5)}$$ Now, we can cancel out the common factor \((x-5)\): $$\lim_{x \rightarrow 5^{+}} \frac{1}{x+5}$$ As \(x\) approaches \(5\) from the right, the limit becomes: $$\lim_{x \rightarrow 5^{+}} \frac{1}{x+5} = \frac{1}{10}$$

Step 4: Case 2 - x < 5

For the case where \(x < 5\), we have: $$\lim_{x \rightarrow 5^{-}} \frac{-(x-5)}{x^{2}-25}$$ Factor the denominator to get: $$\lim_{x \rightarrow 5^{-}} \frac{-(x-5)}{(x-5)(x+5)}$$ Now, we can cancel out the common factor \(-(x-5)\): $$\lim_{x \rightarrow 5^{-}} \frac{-1}{x+5}$$ As \(x\) approaches \(5\) from the left, the limit becomes: $$\lim_{x \rightarrow 5^{-}} \frac{-1}{x+5} = \frac{-1}{10}$$

Step 5: Determine the Overall Limit

Since the limit exists and is the same for both cases, x > 5 and x < 5, we can conclude that the limit of the function as x approaches 5 is: $$\lim_{x \rightarrow 5} \frac{|x-5|}{x^{2}-25} = \frac{1}{10}$$

Key Concepts

Absolute Value Properties

Understanding absolute value properties is crucial to solving problems involving limits. The absolute value of a number, denoted by \(|x|\), represents the distance of that number from zero on the number line, regardless of its direction. This results in two distinct cases:

• If the number inside the absolute value is positive or zero, the absolute value is simply the number itself.
• If the number inside the absolute value is negative, the absolute value mirrors the number, turning it positive by reversing its sign.

For the function \(|x-5|\), this means:- When \(x > 5\), \(|x-5| = x - 5\), because the expression is positive.- When \(x < 5\), \(|x-5| = -(x - 5)\), as the expression is negative, thus requiring a sign change.Breaking down absolute value into two linear conditions helps simplify and understand the analysis of limits, particularly in cases where the input approaches a critical value.

Factoring Polynomials

Factoring polynomials is an essential algebraic skill, especially when simplifying expressions as limits are evaluated. It involves expressing a polynomial as a product of its factors, identifying common factors or using methods like grouping.
In the context of the provided exercise, the denominator \(x^2 - 25\) can be factored using the difference of squares identity, \(a^2 - b^2 = (a-b)(a+b)\). Thus, \(x^2 - 25\) becomes \((x-5)(x+5)\).
This factoring is significant because it allows the cancellation of the \(x-5\) term in the numerator, simplifying the expression.
After cancellation, the expressions become simpler: \((x-5)/(x-5) = 1\), leaving us with the simplified form for further calculations when evaluating limits. Understanding and applying factoring polynomials correctly ensures such simplifications are accurate and aids in finding the limits effectively.

One-Sided Limits

One-sided limits focus on evaluating a function as the variable approaches a particular value from one side - either from the left or the right. This involves:

• \(\lim_{x \rightarrow c^+} f(x)\): Approaching \(c\) from the right (values greater than \(c\)).
• \(\lim_{x \rightarrow c^-} f(x)\): Approaching \(c\) from the left (values less than \(c\)).

In the context of this exercise, a one-sided limit analysis is crucial due to the absolute value expression. By splitting the limit at \(x = 5\) into two parts, we handle the cases where \(x\) approaches 5 from either side:- \(\lim_{x \rightarrow 5^+} \frac{x-5}{(x-5)(x+5)}\), where \(x \rightarrow 5\) from the right, simplifies to \(\frac{1}{10}\).- \(\lim_{x \rightarrow 5^-} \frac{-(x-5)}{(x-5)(x+5)}\), where \(x \rightarrow 5\) from the left, simplifies to \(\frac{-1}{10}\).Understanding one-sided limits helps evaluate discontinuities and provides detailed insight into the behavior of functions near specific points.