Problem 78 - Calculus and its applications

Question

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 3}\left(\frac{x^{3}-x-24}{x^{2}-9}\right) $$

Step-by-Step Solution

Answer

The limit is \(\frac{13}{3}\).

Step 1: Check for Indeterminate Form

Substitute \(x = 3\) into the function \(\frac{x^{3} - x - 24}{x^{2} - 9}\) to check if the limit results in an indeterminate form.ewline\(\text{Numerator: } 3^3 - 3 - 24 = 27 - 3 - 24 = 0\)ewline\(\text{Denominator: } 3^2 - 9 = 9 - 9 = 0\)ewlineBoth numerator and denominator equal zero, indicating a \(\frac{0}{0}\) indeterminate form.

Step 2: Differentiate the Numerator and Denominator Separately

Differentiate the numerator \(x^3 - x - 24\) to get \(3x^2 - 1\).ewlineDifferentiate the denominator \(x^2 - 9\) to get \(2x\).ewlineThus, \(\frac{d}{dx}(x^3 - x - 24) = 3x^2 - 1\) and \(\frac{d}{dx}(x^2 - 9) = 2x\).

Step 3: Substitute Back into the Limit

Use l'Hôpital's Rule and substitute into the new limit. The limit becomes \(\lim_{x \rightarrow 3}\left(\frac{3x^2 - 1}{2x}\right)\).

Step 4: Evaluate the New Limit

Substitute \(x = 3\) into \(\frac{3x^2 - 1}{2x}\):ewline\(\text{Numerator: } 3(3)^2 - 1 = 3(9) - 1 = 27 - 1 = 26\)ewline\(\text{Denominator: } 2(3) = 6\)ewlineSo, \(\lim_{x \rightarrow 3}\left(\frac{3x^2 - 1}{2x}\right) = \frac{26}{6} = \frac{13}{3}\).

Key Concepts

Indeterminate Forms

In calculus, when we deal with limits, we often come across expressions that are not immediately clear. Such expressions are known as "indeterminate forms." They typically occur when trying to evaluate limits that result in forms like \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). These forms are labeled indeterminate because they don't provide enough information to determine the actual limit value immediately.

When we encounter an indeterminate form, it signifies that the simplistic direct substitution or evaluation will not work. Instead, it signals the need for more advanced techniques to find the limit. In the context of the original exercise, we observe the form \( \frac{0}{0} \) by substituting the values directly into the function \( \frac{x^3 - x - 24}{x^2 - 9} \) at \( x = 3 \).

This type of form is why l'Hôpital's Rule becomes fundamentally useful, providing a method to resolve the expression using differentiation rather than direct substitution.

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. It is instrumental in analyzing and determining the rate at which a function changes. Differentiation transforms a function into its derivative, providing insight into the slope of the tangent line or the rate of change at any given point.

In the context of indeterminate forms, particularly when applying l'Hôpital's Rule, differentiation serves as a tool to simplify and solve the complex limits we encounter. In the given exercise, to address the indeterminate form \( \frac{0}{0} \), we differentiate both the numerator and the denominator separately:

• The numerator \( x^3 - x - 24 \) is differentiated to \( 3x^2 - 1 \).
• The denominator \( x^2 - 9 \) is differentiated to \( 2x \).

By doing this, we transform the original problem into a simpler one where we can easily evaluate using new functions formed from their derivatives.

Limits in Calculus

Limits are a fundamental part of calculus and are used to explore the behavior of functions as they approach particular points. Limits allow us to deal with situations where direct evaluation is not possible, particularly in cases dealing with continuous functions or infinite behaviors.

In the original exercise, the application of limits comes through finding the limit of the expression \( \frac{x^3 - x - 24}{x^2 - 9} \) as \( x \) approaches 3. Normally, you'd plug in the value directly, but due to the indeterminate form \( \frac{0}{0} \), this approach fails.

To resolve this, after differentiating using l'Hôpital's Rule, the new limit becomes \( \lim_{x \rightarrow 3}\left(\frac{3x^2 - 1}{2x}\right) \). Substituting \( x = 3 \) into this revised expression allows for a straightforward evaluation, giving a finite limit of \( \frac{13}{3} \). This showcases the power and precision of using limits in calculus, especially when combined with different techniques like differentiation to solve otherwise intractable problems.