Problem 28 - Contemporary Precalculus

Question

Use the Infinite Limit Theorem and the properties of limits to find the limit. $$\lim _{x \rightarrow-\infty} \frac{x}{\sqrt{x^{2}-1}}$$

Step-by-Step Solution

Answer

Answer: The limit of the given function as x tends to negative infinity does not exist.

Step 1: Basic simplification

Let's first simplify the expression by dividing both numerator and denominator by x: $$\lim _{x \rightarrow-\infty} \frac{x}{\sqrt{x^{2}-1}} = \lim _{x \rightarrow-\infty} \frac{1}{\sqrt{\frac{x^2-1}{x^2}}}$$

Step 2: Further simplification

Now, simplify the expression further by applying the fact that: $$\lim_{x\to \infty} \frac{f(x)}{g(x)} = \lim_{x\to \infty} \frac{f(x)}{g(x)}\cdot\frac{1}{x}\cdot x = \frac{\lim_{x\to \infty} f(x)\cdot \frac{1}{x}}{\lim_{x\to \infty} g(x)\cdot \frac{1}{x}}.$$ So, we have: $$\lim _{x \rightarrow-\infty} \frac{1}{\sqrt{\frac{x^2-1}{x^2}}} = \lim _{x \rightarrow-\infty} \frac{\frac{1}{x}}{\sqrt{\frac{x^2-1}{x^2}}\cdot \frac{1}{x}}$$

Step 3: Apply limit

Now, let's apply the limit to the simplified expression: $$\lim _{x \rightarrow-\infty} \frac{\frac{1}{x}}{\sqrt{\frac{x^2-1}{x^2}}\cdot \frac{1}{x}} = \frac{\lim_{x\rightarrow -\infty} \frac{1}{x}}{\lim_{x\rightarrow -\infty} \sqrt{\frac{x^2-1}{x^2}}\cdot \frac{1}{x}}$$

Step 4: Evaluate limits

Evaluating the limits in the numerator and denominator: $$\lim_{x\rightarrow -\infty} \frac{1}{x} = 0$$ And, $$\lim_{x\rightarrow -\infty} \sqrt{\frac{x^2-1}{x^2}}\cdot \frac{1}{x} = \lim_{x\rightarrow -\infty} \sqrt{1-\frac{1}{x^2}}\cdot \lim_{x\rightarrow -\infty} \frac{1}{x}$$ Since \(\lim_{x\rightarrow -\infty} \sqrt{1-\frac{1}{x^2}} = 1\) and \(\lim_{x\rightarrow -\infty} \frac{1}{x} = 0,\) their product is also 0: $$\lim_{x\rightarrow -\infty} \sqrt{\frac{x^2-1}{x^2}}\cdot \frac{1}{x} = 1 \cdot 0 = 0$$

Step 5: Final result

Now, we can conclude with our final result: $$\frac{\lim_{x\rightarrow -\infty} \frac{1}{x}}{\lim_{x\rightarrow -\infty} \sqrt{\frac{x^2-1}{x^2}}\cdot \frac{1}{x}} = \frac{0}{0}$$ However, this result is undefined. So, the limit of the given function as x tends to negative infinity does not exist.

Key Concepts

Properties of Limits

When we discuss the properties of limits, we are referring to the set of rules that help us find the limits of functions efficiently. Understanding these properties is crucial for an in-depth comprehension of limits. Here are a few key properties:

• Sum Property: If the limit of two functions, say \(f(x)\) and \(g(x)\), exists as \(x\) approaches a point, then the limit of their sum will equal the sum of their individual limits: \( \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \).
• Product Property: If the limits of \(f(x)\) and \(g(x)\) as \(x\) approaches a point both exist, then the limit of the product is the product of their limits: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \).
• Quotient Property: If the limits of \(f(x)\) and \(g(x)\) exist, and \(\lim_{x \to a} g(x) eq 0\), then the limit of the quotient is the quotient of the limits: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \).

These properties help simplify complex expressions and evaluate limits more effectively, guiding us in exercises like finding the limit as \(x\) approaches negative infinity.

Evaluating Limits

Evaluating limits involves determining the value a function approaches as the input approaches a specific number or infinity. In this problem, we're examining how the function behaves as \(x\) goes to negative infinity. Here’s a simple way to handle such evaluations:

• Identify dominant terms in the numerator and denominator. These are often terms with the highest power.
• Simplify the expression using algebraic manipulation, ensuring the form is manageable.
• Apply limits to each component separately using properties of limits.

By breaking down the original function to its simplest form, we made it easier to decide if the limit exists and, if so, to determine what it is. This process highlights the importance of each algebraic step in evaluating the given limits as in the exercise.

Limit Does Not Exist

Sometimes, while evaluating limits, you encounter a situation where the limit does not exist. This can happen for several reasons:

• The function approaches different values from the left and right side.
• The function oscillates between values without settling on a single one.
• The expression results in an indeterminate form like \( \frac{0}{0} \).

In the given problem, we simplified the expression to end up with an indeterminate form \( \frac{0}{0} \). While indeterminate forms suggest that the limit might exist, they also signal the need for further evaluation or alternative approaches like L'Hôpital's Rule or algebraic manipulation. Here, we concluded that the expression remains undefined, meaning the limit does not exist for this particular problem.

Negative Infinity

When dealing with limits as \(x\) approaches negative infinity, we focus on how the terms behave as their values grow significantly large negatively. This can affect both the numerator and denominator in expressions:

• Terms with the highest power in \(x\) often dominate the behavior.
• Negative values can affect the sign and limit's behavior, especially with even-powered expressions resulting in positive outcomes.

In the exercise, taking the limit as \(x\) goes to negative infinity required consideration of these negative and positive interactions, particularly after simplifying. This assists in accurately determining whether the function approaches infinity, zero, or another state. Hence, evaluating behavior in the realm of negative infinity is vital to reaching conclusions for limits effectively. By understanding these interactions, students can master this abstract concept and handle similar problems more confidently.