Problem 84 - 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. $$ \lim _{x \rightarrow-\infty}\left(\frac{3 x^{3}+x+11}{6 x^{3}+x+2}\right) $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Check Initial Form

First, we substitute \( x = -\infty \) directly into \( \frac{3x^3 + x + 11}{6x^3 + x + 2} \) to determine the expression's form. As \( x \to -\infty \), the terms dominated by \( x^3 \) are \( 3x^3 \) and \( 6x^3 \). Thus, we get \( \frac{3(-\infty)^3}{6(-\infty)^3} = \frac{-\infty}{-\infty} \), an indeterminate form.

Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{-\infty}{-\infty} \), we can apply L'Hôpital's Rule. Differentiate the numerator and the denominator separately: \( (3x^3 + x + 11)' = 9x^2 + 1 \) and \( (6x^3 + x + 2)' = 18x^2 + 1 \). Then, evaluate the limit of these derivatives as \( x \to -\infty \).

Step 3: Evaluate the New Limit

Calculate \( \lim_{x \to -\infty} \frac{9x^2 + 1}{18x^2 + 1} \). As \( x \to -\infty \), the dominant terms are \( 9x^2 \) in the numerator and \( 18x^2 \) in the denominator. Thus, \( \lim_{x \to -\infty} \frac{9x^2 + 1}{18x^2 + 1} = \lim_{x \to -\infty} \frac{9x^2}{18x^2} = \frac{9}{18} = \frac{1}{2} \).

Step 4: Conclusion

Since the new limit, after applying L'Hôpital's Rule, is \( \frac{1}{2} \), this is also the limit of the original function as \( x \to -\infty \). Therefore, \( \lim_{x \to -\infty} \frac{3x^3 + x + 11}{6x^3 + x + 2} = \frac{1}{2} \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms occur when direct substitution in a limit gives an unclear or ambiguous result. Common examples of this are \( \frac{0}{0} \) and \( \frac{\pm \infty}{\pm \infty} \).

These forms do not provide specific values, and different methods need to be employed to find the actual limit. Here are some basic types of indeterminate forms:

• \(\frac{0}{0}\) arises when both the numerator and denominator approach zero.
• \(\frac{\infty}{\infty}\) happens when both approach infinity.

When faced with these forms, you can't simply plug the values into the expression. Instead, you need strategies like L'Hôpital's Rule to determine the limit. Understanding indeterminate forms is crucial because they help identify when straightforward arithmetic isn't enough to solve a problem.

Differentiation

Differentiation is a fundamental process in calculus used to find the rate at which one quantity changes with respect to another. In the context of L'Hôpital's Rule, differentiation allows us to simplify the evaluation of limits that initially present as indeterminate forms.

Differentiation involves:

• Identifying the function that needs to be differentiated, such as the numerator and denominator of a fraction.
• Applying derivative rules—like the power rule, product rule, or chain rule—to find the derivative of these functions.

For example, differentiating the numerator \(3x^3 + x + 11\) yields \(9x^2 + 1\), and differentiating the denominator \(6x^3 + x + 2\) yields \(18x^2 + 1\).

After differentiation, you can re-evaluate the limit. This application of derivatives helps transform the original problem into a simpler one, making it easier to solve.

Limit Evaluation

Limit evaluation is the process of finding the value that a function approaches as the input approaches a particular point. This concept is at the heart of L'Hôpital's Rule, a technique used to resolve limits that lead to indeterminate forms.

To evaluate limits effectively:

• Begin by attempting to substitute the value to see if an indeterminate form arises.
• If so, apply L'Hôpital's Rule by differentiating the numerator and denominator separately.
• Reassess the limit of the new functions. In many cases, the dominating terms simplify the evaluation.

For instance, after applying L'Hôpital's Rule to our original problem, we examine the limit \(\lim_{x \to -\infty} \frac{9x^2 + 1}{18x^2 + 1}\).
By focusing on the dominant terms \(9x^2\) and \(18x^2\), we simplify the limit to \(\frac{9}{18} = \frac{1}{2}\).

This demonstrates how evaluating limits transforms complex expressions into straightforward results.