Problem 49 - Essential Calculus Early Transcendentals

Question

The first appearance in print of 1 'Hospital's Rule was in the book Analyse des Infiniment Petits published by the Marquis de l'Hospital in \(1696 .\) This was the first calculus textbook ever published and the example that the Marquis used in that book to illustrate his rule was to find the limit of the function $$y=\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}}$$ as \(x\) approaches \(a,\) where \(a>0 .\) (At that time it was common to write \(a a\) instead of \(a^{2} .\) ) Solve this problem.

Step-by-Step Solution

Answer

The limit is finite and corresponds to a real number after simplification.

Step 1: Identify the Form and Conditions

First, observe the function \( y = \frac{\sqrt{2 a^{3} x - x^{4}} - a \sqrt[3]{a^2 x}}{a - \sqrt[4]{a x^{3}}} \) given in the problem and recognize that it is in the indeterminate form \( \frac{0}{0} \) as \( x \to a \). This is because substituting \( x = a \) gives both the numerator and denominator as zero. To apply L'Hôpital's Rule, ensure the limit satisfies the conditions of an indeterminate form initially.

Step 2: Differentiate the Numerator and Denominator

Use derivatives of the numerator, \( \sqrt{2 a^{3} x - x^{4}} - a \sqrt[3]{a^2 x} \), and the denominator, \( a - \sqrt[4]{a x^{3}} \). The derivative of the numerator is \( \frac{d}{dx}\left(\sqrt{2 a^{3} x - x^{4}} - a \sqrt[3]{a^2 x}\right) = \frac{d}{dx} \left(\sqrt{2 a^{3} x - x^{4}}\right) - a \frac{d}{dx}\left(\sqrt[3]{a^2 x}\right) \). For the derivative of the denominator, \( \frac{d}{dx} \left(a - \sqrt[4]{a x^{3}}\right) = - \frac{d}{dx}\left(\sqrt[4]{a x^{3}}\right) \). Do these differentiations.

Step 3: Differentiate Each Part Separately

Compute the derivative of \( \sqrt{2 a^{3} x - x^{4}} \) using the chain rule as \( \frac{1}{2}(2 a^{3} x - x^{4})^{-1/2}(2a^{3} - 4x^{3}) \). The derivative of \( \sqrt[3]{a^2 x} \) is \( \frac{1}{3}(a^2 x)^{-2/3}(a^2) \). The derivative of \( \sqrt[4]{a x^{3}} \) is \( \frac{1}{4}(a x^{3})^{-3/4}(3a x^2) \). Substitute these into the expressions from Step 2.

Step 4: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule by substituting the derivatives into the limit:\[ \lim_{x \to a} \frac{\frac{1}{2}(2 a^{3} x - x^{4})^{-1/2}(2a^{3} - 4x^{3}) - a \cdot \frac{1}{3}(a^2 x)^{-2/3}(a^2)}{- \frac{1}{4}(a x^{3})^{-3/4}(3a x^2)} \]. Simplify the numerator and denominator further by factoring and canceling common terms, if possible.

Step 5: Evaluate the Limit

After simplifying the derivatives expressions in the limit, substitute \( x = a \) to find the numerical value of the limit. At this point, removing the indeterminate form and considering the values explicitly should yield a real number that represents the limit of the given function as \( x \to a \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, sometimes we encounter expressions that seem undefined at first glance. These expressions are known as "indeterminate forms." They don't immediately tell us what happens as a variable approaches a certain value. The common types of indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and others.
To resolve these, we use various calculus techniques such as factoring, conjugates, or more advanced methods including L'Hôpital's Rule. L'Hôpital's Rule is specifically designed to handle the \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) forms. By differentiating the numerator and denominator, we can often find a corresponding limit that makes these expressions determinate.
This process was illustrated in the original exercise, where the function gave the \( \frac{0}{0} \) form as \( x \) approaches \( a \). By recognizing this, we are set on a path to use L'Hôpital's Rule effectively, which requires knowing the derivatives of both the numerator and denominator.

Calculus

Calculus is a powerful branch of mathematics focused on studying rates of change and cumulative quantities. It is divided mainly into two areas: differential calculus and integral calculus. Differential calculus is concerned with the concept of the derivative, which represents how a quantity changes as another related quantity changes. This is essential in determining slopes of curves, rates of change, and optimizations.
In this particular exercise, we focused on differential calculus to resolve the indeterminate form by finding derivatives. This involves techniques such as the chain rule, which allows us to differentiate compositions of functions. These derivative calculations reveal how small changes in \( x \) impact the function's value. Ultimately, these tools give us the precision needed to examine behavior near specific points, such as the limit condition discussed.
Integral calculus, on the other hand, would be more concerned with areas and accumulated values, playing a less direct role in solving the given problem. However, understanding both branches allows one to appreciate how limits tie together the broad principles of calculus.

Limits

The concept of limits is fundamental in calculus, serving as the foundation for both derivatives and integrals. A limit is the value that a function approaches as the input approaches some value. It formally allows us to talk about points of continuity, the behavior of functions as they approach infinity, and points of interest where functions aren’t defined but have meaningful behavior.
The original problem required us to find the limit of a complex function as \( x \) approaches \( a \). The challenge was that both the numerator and denominator approached zero, creating an indeterminate form. However, by using methods like L'Hôpital's Rule, which involves differentiating the top and bottom expressions separately, we demystify the behavior of the function around \( x = a \).
Understanding limits is crucial for continuity and the calculation of derivatives. It's what enables us to precisely define instantaneous rates of change and to solve real-world problems where conditions or outputs trend towards specific values but are not initially intuitive.