Problem 66 - Calculus for Biology and Medicine

Question

When l'Hospital introduced indeterminate limits in his textbook, his first example was $$ \lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}} $$ where \(a\) is a positive constant. (This example was communicated to him by Bernoulli.) Show that this limit is equal to \((16 / 9) a\).

Step-by-Step Solution

Answer

The limit is \(\frac{16}{9}a\).

Step 1: Identify the Indeterminate Form

Plug in the value of \(x = a\) into the limit expression to check the form. Substitute \(x = a\) in the numerator and denominator: \(\sqrt{2a^4 - a^4} - a\sqrt[3]{a^3}\) and \(a - (a^{4/4}) = a - a = 0\). Both the numerator and denominator give 0, indicating an indeterminate form \(\frac{0}{0}\).

Step 2: Apply L'Hôpital's Rule

Since the expression is in \(\frac{0}{0}\) form, apply L'Hôpital's Rule which involves differentiating the numerator and denominator with respect to \(x\). Compute the derivative of the numerator: \(\frac{d}{dx}(\sqrt{2a^3x - x^4} - a\sqrt[3]{a^2 x})\). Compute the derivative of the denominator: \(\frac{d}{dx}(a - \sqrt[4]{a x^3})\).

Step 3: Differentiate the Numerator

The numerator is \(\sqrt{2a^3x - x^4} - a\sqrt[3]{a^2 x}\). The derivative of \(\sqrt{2a^3x - x^4}\) is \(\frac{1}{2\sqrt{2a^3x - x^4}}(2a^3 - 4x^3)\) by applying the chain rule. The derivative of \(-a\sqrt[3]{a^2x}\) is \(-a \cdot \frac{1}{3}(a^2x)^{-2/3} \cdot a^2\).

Step 4: Differentiate the Denominator

The denominator is \(a - \sqrt[4]{ax^3}\). The derivative is 0 for \(a\). The derivative of \(\sqrt[4]{ax^3}\) is \(-\frac{1}{4}(ax^3)^{-3/4}\cdot 3ax^2\).

Step 5: Simplify the Derivatives

Substitute the derivatives calculated in Steps 3 and 4 back into the expression for the limit: \[\lim_{x \to a} \frac{\frac{a^3 - 2x^3}{\sqrt{2a^3x - x^4}} - \frac{2a^3}{3(a^2x)^{2/3}}}{-\frac{3ax^2}{4(ax^3)^{3/4}}}\] Simplify both the numerator and the denominator.

Step 6: Evaluate the Limit

Substitute \(x = a\) into the simplified derivatives, and calculate the limit value: \[\lim_{x \to a} \frac{a^3 - 2a^3}{\sqrt{a^4}} - \frac{2a^3}{3(a^3)^{2/3}}\right)/\frac{3a^3}{4(a^4)^{3/4}}}\]. After simplification, the result should equal \(\frac{16}{9}a\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool in calculus for finding limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule gets its name from the French mathematician Guillaume de l'Hôpital, who formalized it in his work. The basic idea is that if you encounter an indeterminate form, you can differentiate the numerator and the denominator separately and then take the limit again. This can often simplify the process of finding the limit significantly.

Here’s a quick refresher:

• Check the form of your limit. If both the numerator and denominator result in 0, you're facing a \(\frac{0}{0}\) indeterminate form.
• Then, differentiate the numerator and the denominator separately with respect to the variable, in our case, \(x\).
• Finally, take the limit of this new fraction. If the result is still in indeterminate form, you can apply l'Hôpital's Rule repeatedly.

In the exercise provided, the indeterminate form \(\frac{0}{0}\) was resolved using this rule to compute a more manageable expression whose limit was easier to evaluate.

Indeterminate Forms

Indeterminate forms are expressions in calculus that don't neatly result in a defined value upon initial evaluation, especially when computing limits. These expressions can trick you into thinking a limit doesn’t exist, even when it might. Common indeterminate forms are \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), \(1^\infty\), \(0^0\), and \(\infty - \infty\).

Understanding indeterminate forms is crucial for correctly analyzing functions and their limits. Here's why:

• They tell us that just substituting a value into an equation isn't enough to find the limit.
• You need additional methods, like L'Hôpital's Rule or algebraic manipulation, to evaluate them.

In this exercise, substituting \(x = a\) initially resulted in \(\frac{0}{0}\), prompting the need for alternative methods to solve for the limit. Recognizing an indeterminate form is often the first step in correctly applying L'Hôpital's Rule.

Limits in Mathematics

In mathematics, limits help us understand how a function behaves as it approaches a specific point. They are a cornerstone of calculus, serving as the foundation for defining derivatives and integrals.

Calculating limits involves the following:

• Asking what value the function approaches as \(x\) gets arbitrarily close to a number, say \(a\).
• Understanding that limits do not necessarily represent the value of the function at that point, only what the function heads towards.
• Being aware that limits can exist even if the function doesn't reach that specific point.

In the given exercise, the expression involved was examined as \(x\) approached \(a\). By using calculus techniques such as differentiation and rules like L'Hôpital's, we unearthed the limit, \(\frac{16}{9}a\). Understanding limits provides insight into the continuity and behavior of functions across their domains.