Problem 11 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow \infty} \frac{5 x^{3}-2 x}{7 x^{3}+3}$$

Step-by-Step Solution

Answer

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

Step 1: Confirm Indeterminate Form

First, substitute \( x = \infty \) into the expression \( \frac{5x^3 - 2x}{7x^3 + 3} \). Both the numerator and denominator tend towards infinity, which indicates the form \( \frac{\infty}{\infty} \). This confirms that we can use L'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator

Apply L'Hôpital's Rule by differentiating the numerator \(5x^3 - 2x\) and the denominator \(7x^3 + 3\). The derivative of the numerator is:\[ \frac{d}{dx}(5x^3 - 2x) = 15x^2 - 2 \]The derivative of the denominator is:\[ \frac{d}{dx}(7x^3 + 3) = 21x^2 \]

Step 3: Apply L'Hôpital's Rule

Substitute the derivatives back into the limit:\[ \lim_{x \to \infty} \frac{15x^2 - 2}{21x^2} \]

Step 4: Simplify the Expression

Simplify the expression by dividing each term in the numerator by \( x^2 \).\[ \lim_{x \to \infty} \frac{15x^2 - 2}{21x^2} = \lim_{x \to \infty} \frac{15 - \frac{2}{x^2}}{21} \]

Step 5: Evaluate the Simplified Limit

As \(x \to \infty\), the term \(\frac{2}{x^2}\) approaches 0.Hence, \( \lim_{x \to \infty} \frac{15 - \frac{2}{x^2}}{21} = \frac{15}{21} = \frac{5}{7} \).

Key Concepts

Limits of Functions

Understanding the limits of functions is an essential step in calculus. Limits help us to understand how functions behave as variables approach a specific value. In simpler terms, it tells us what a function looks like ‘near’ some point, even if it doesn’t exactly reach that point.

When learning about limits, remember:

• Substituting the value, the variable approaches, into the function can give direct results if the function is continuous at that point.
• For functions approaching infinity (\(x \rightarrow \infty\)], the idea is to observe how the function behaves as it grows larger or smaller without bound.
• Limits can help in understanding infinitely large and small values, aiding in calculations like derivatives.

In cases where direct substitution results in a meaningful value, that is the limit. If not, or in cases of indeterminate forms, we use tools like factoring, simplifying, or even L'Hôpital's Rule for finding the limit.

Indeterminate Forms

Indeterminate Forms occur when a straightforward substitute does not yield a clear answer for a limit. You encounter these forms when both the numerator and denominator of a function go towards 0 (\(0/0\)) or infinity (\(\infty/\infty\)). Such forms do not tell you what the limit exactly is, so further analysis is needed.

Here’s what to consider with indeterminate forms:

• Common indeterminate forms include: - \(\frac{0}{0}\): Often arises from polynomials and rational functions. - \(\frac{\infty}{\infty}\): Occurs in limits involving asymptotic behavior.
• L'Hôpital's Rule is a powerful method to resolve many indeterminate forms by differentiation.
• Look for opportunities to simplify or factor functions to resolve a clearer limit before resorting to advanced techniques.

In the exercise, the form \(\frac{\infty}{\infty}\) suggests deploying L'Hôpital's Rule to make the problem solvable.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. They play a crucial role in calculus, especially in determining limits and behaviors at boundaries.

Understanding rational functions includes:

• Checking for zeros of the denominator which indicates vertical asymptotes.
• Understanding that the degree of polynomials in the numerator and denominator affects the horizontal asymptote and end behavior.
• Simplifying expressions by factoring can help in finding zeros and simplifying limits.

In the original exercise, we dealt with a rational function of high degree. After simplifying using L'Hôpital's Rule, a clear limit was found, demonstrating that as x approaches infinity, expressions simplify based on the leading coefficients, resulting in a more manageable and intelligible function.