Problem 3 - University Calculus: Early Transcendentals

Question

Use I'Hópital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. $$\lim _{x \rightarrow \infty} \frac{5 x^{2}-3 x}{7 x^{2}+1}$$

Step-by-Step Solution

Answer

The limit is \( \frac{5}{7} \).

Step 1: Identify Indeterminate Form

First, let's check if the limit \( \lim_{x \to \infty} \frac{5x^2 - 3x}{7x^2 + 1} \) presents an indeterminate form. As \( x \to \infty \), both the numerator and denominator approach infinity, creating an \( \frac{\infty}{\infty} \) form suitable for L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, when a limit is in the form of \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately and then re-evaluate the limit. First, differentiate the numerator: \( \frac{d}{dx}(5x^2 - 3x) = 10x - 3 \). Then, differentiate the denominator: \( \frac{d}{dx}(7x^2 + 1) = 14x \). This transforms the limit to \( \lim_{x \to \infty} \frac{10x - 3}{14x} \).

Step 3: Simplify and Evaluate New Limit

Simplify the expression \( \frac{10x - 3}{14x} \) by dividing both the numerator and the denominator by \( x \) to get \( \frac{10 - \frac{3}{x}}{14} \). As \( x \to \infty, \frac{3}{x} \to 0 \), leaving the simplified expression as \( \frac{10}{14} = \frac{5}{7} \). Thus, \( \lim_{x \to \infty} \frac{10x - 3}{14x} = \frac{5}{7} \).

Step 4: Alternative Method - Leading Term Comparison

Without using L'Hôpital, observe the original limit and focus on the highest degree terms in both the numerator and denominator: \( \frac{5x^2}{7x^2} \). Since the other terms become negligible as \( x \to \infty \), the limit simplifies to \( \lim_{x \to \infty} \frac{5x^2}{7x^2} = \frac{5}{7} \) directly.

Key Concepts

Indeterminate Form

When evaluating limits, sometimes we encounter expressions that do not clearly indicate a well-defined finite or infinite value. These are known as indeterminate forms. One common type is the \( \frac{\infty}{\infty} \) form. This happens when both the numerator and denominator of a fraction tend towards infinity as the variable approaches a certain value, like in our given expression: \[ \lim _{x \rightarrow \infty} \frac{5 x^{2}-3 x}{7 x^{2}+1} \] In this case, both the top and bottom parts approach infinity when \( x \to \infty \). This situation is ambiguous because infinity divided by infinity doesn't have a clear result. To handle it, we usually apply special techniques, like L'Hôpital's Rule, which are designed to help resolve these indeterminate forms.

Limits

The concept of limits is foundational in calculus. Limits help us understand the behavior of a function as it approaches a certain point or infinity. In our example, we want to find out what value the function approaches as \( x \to \infty \). Consider the limit notation: \[ \lim _{x \rightarrow \infty} \frac{5 x^{2}-3 x}{7 x^{2}+1} \] This tells us to look at the function \( \frac{5 x^{2}-3 x}{7 x^{2}+1} \) as \( x \) increases indefinitely. The challenge is to determine how the function behaves and reach a conclusion about its limiting value. By using techniques such as L'Hôpital's Rule, we can simplify complex expressions and find these limits more efficiently. This is especially useful in calculus where precise values are necessary for further analysis.

Leading Term Comparison

When evaluating limits, especially as \( x \to \infty \), a practical approach is the leading term comparison. This involves focusing on the highest degree terms in both the numerator and the denominator of a rational function. Consider the expression: \[ \frac{5x^2 - 3x}{7x^2 + 1} \] Here, the leading terms are \( 5x^2 \) in the numerator and \( 7x^2 \) in the denominator. As \( x \) becomes very large, the other terms (\(-3x\) and \(1\)) become negligible. So, the expression simplifies to: \[ \lim_{x \to \infty} \frac{5x^2}{7x^2} = \frac{5}{7} \] This method saves time and provides a direct way to evaluate limits without additional computation. It emphasizes that in dominant-degree polynomial functions, only the leading terms govern the behavior as \( x \to \infty \).