Problem 80 - 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. $$ \lim _{x \rightarrow-4}\left(\frac{x^{2}+x-12}{x+4}\right) $$

Step-by-Step Solution

Answer

The limit is -7.

Step 1: Check for Indeterminate Form

Evaluate the expression by substituting the limit value, \( x = -4 \), into the fraction.\[ f(-4) = (-4)^2 + (-4) - 12 = 16 - 4 - 12 = 0 \] \[ g(-4) = -4 + 4 = 0 \] This results in \( \frac{0}{0} \), which is an indeterminate form, making the application of L'Hôpital's Rule appropriate.

Step 2: Differentiate the Numerator and Denominator

Find the derivative of the numerator \( f(x) = x^2 + x - 12 \) and the denominator \( g(x) = x + 4 \). \[ f'(x) = \frac{d}{dx}(x^2 + x - 12) = 2x + 1 \] \[ g'(x) = \frac{d}{dx}(x + 4) = 1 \]

Step 3: Apply L'Hôpital's Rule

Use L'Hôpital's Rule to find the limit of the fraction of the derivatives. \[ \lim_{x \to -4} \frac{x^2 + x - 12}{x + 4} = \lim_{x \to -4} \frac{f'(x)}{g'(x)} = \lim_{x \to -4} \frac{2x + 1}{1} \]

Step 4: Evaluate the Limit

Substitute \( x = -4 \) into the derivative limit expression. \[ \lim_{x \to -4} (2x + 1) = 2(-4) + 1 = -8 + 1 = -7 \] The limit is \( -7 \).

Key Concepts

Indeterminate Forms

Indeterminate forms occur when evaluating certain limit expressions in calculus, particularly in the forms of \( \frac{0}{0} \) and \( \frac{\pm \infty}{\pm \infty} \). These forms do not initially provide enough information to determine a limit directly. They are called "indeterminate" because the limit could potentially evaluate to a range of different values, depending on the specific function involved.
To handle indeterminate forms, one useful technique is l'Hôpital's Rule, which uses differentiation to find the limit. This concept advises us that if a direct substitution into \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in an indeterminate form like \( \frac{0}{0} \), then under certain conditions, the limit of the derivatives \( \frac{f'(x)}{g'(x)} \) can be calculated instead. This approach helps to refine our understanding and resolve ambiguities present in the original form. Thus, l'Hôpital's Rule acts as a tool to "decompose" the complexity.Understanding these forms and applying l'Hôpital's Rule correctly is crucial to advancing in calculus. It allows students to solve problems involving limits that are otherwise difficult to evaluate using direct substitution.

Differentiation

Differentiation is a fundamental concept in calculus, representing the process of calculating a derivative. The derivative is essentially the rate at which a function is changing at any given point, giving insight into its behavior and trends. When applying l'Hôpital's Rule, the power of differentiation is particularly highlighted.
In our example, the original limit resulted in \( \frac{0}{0} \), an indeterminate form. Here, we differentiate both the numerator and the denominator separately.

• For the numerator \( f(x) = x^2 + x - 12 \), we calculate its derivative to obtain \( f'(x) = 2x + 1 \).
• Similarly, the denominator \( g(x) = x + 4 \) yields a constant derivative of \( g'(x) = 1 \).

By differentiating component-wise, we transform the initial indeterminate problem into something more manageable.
This ability to simplify complex equations through differentiation is what makes it a cornerstone of calculus, enhancing our toolkit when dealing with challenging problems and atypical limit expressions.

Limits in Calculus

Limits are a core component of calculus, foundational for defining derivatives and integrals. A limit helps you understand the behavior of a function as it approaches a particular point. In calculus, we often explore the limit of a function as the variable approaches a specific value (such as \( x \to a \)).
In the context of l'Hôpital's Rule, evaluating limits is crucial when solving indeterminate forms. It allows us to probe functions' behaviors deeply around challenging points. Consider our example, where we had the expression \( \lim_{x \to -4} \frac{x^2 + x - 12}{x + 4} \). Initially, substitution of \( x = -4 \) resulted in the \( \frac{0}{0} \) indeterminate form. Through differentiation and re-evaluation of the limit of the new function, \( \frac{2x + 1}{1} \), we precisely found the limit of \(-7\).
Understanding limits is a step towards mastering continuity and understanding more complex topics within calculus. They give us the "big picture" view of how functions behave in limiting cases, equipping us with the analytical skills to solve real-world problems effectively and creatively.