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Evaluate one of the limits I'Hôpital used in his own textbook in about 1700 : $$\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}, \text { where } a \text { is a real number. }$$

Short Answer

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Question: Evaluate the limit \(\displaystyle\lim_{x \rightarrow a}\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2}x}}{a-\sqrt[4]{a x^{3}}}\) using l'Hôpital's Rule. Answer: \(\frac{2}{3}\)

Step by step solution

01

Determine the form of the limit

First, we need to determine which form (case 1 or case 2) our limit falls into. Let's plug in x=a and see what happens. $$\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2}x}}{a-\sqrt[4]{a x^{3}}}=\frac{\sqrt{2 a^{4}-a^{4}}-a \sqrt[3]{a^{3}}}{a-\sqrt[4]{a^4}}=\frac{0}{0}$$ The result is of the indeterminate form \(\frac{0}{0}\), which means we can use l'Hôpital's Rule.
02

Apply l'Hôpital's Rule

By applying l'Hôpital's Rule, we can differentiate both the numerator and the denominator with respect to x and then find the limit as x approaches a. \[ \lim _{x \rightarrow a} \frac{\frac{d}{dx}(\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2}x})}{\frac{d}{dx}(a-\sqrt[4]{a x^{3}})} \]
03

Differentiate the numerator and denominator

Now, we'll differentiate the numerator and denominator separately. For the numerator: \[ \frac{d}{dx}(\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2}x}) =\frac{d}{dx}\left(\sqrt{2 a^{3} x-x^{4}}\right) -\frac{d}{dx}(a \sqrt[3]{a^{2}x}) \] Applying chain rule for both terms: $$ =\frac{1}{2}(2 a^{3} x-x^{4})^{-\frac{1}{2}}(6a^2 -4x^3) - a\sqrt[3]{a^2}\cdot\frac{1}{3}(a^{2} x)^{-\frac{2}{3}} $$ Simplifying the expression: $$ =\frac{6a^2-4x^3}{2\sqrt{2a^3x-x^4}} -\frac{a}{3x^{\frac{2}{3}}} $$ For the denominator, we have: $$ \frac{d}{dx}(a-\sqrt[4]{a x^{3}})=-\frac{1}{4}(a x^{3})^{-\frac{3}{4}}(3ax^2) $$ Simplifying the expression: $$ =-\frac{3a}{4x^{\frac{3}{4}}} $$
04

Find the new limit

Now that we've differentiated the numerator and denominator, let's plug them back into the limit and simplify: \[ \lim_{x\rightarrow a} \frac{\frac{6a^2-4x^3}{2\sqrt{2a^3x-x^4}} -\frac{a}{3x^{\frac{2}{3}}}}{-\frac{3a}{4x^{\frac{3}{4}}}} \] Next, cancel out the common factors: \[ \lim_{x\rightarrow a} \frac{\frac{1}{x^{\frac{3}{4}}}(6a^2-4x^3)}{2\sqrt{2a^3x-x^4}} -\frac{1}{x^{\frac{2}{3}}} \cdot x^{\frac{3}{4}} \frac{a}{3} \cdot \frac{4x^{\frac{3}{4}}}{3a} }=\lim_{x\rightarrow a} \frac{\frac{6a^2-4x^3}{2\sqrt{2a^3x-x^4}} -\frac{4}{3}} $$ Let's plug x=a into the simplified expression: \[ \frac{6a^2-4a^3}{2\sqrt{2a^4-a^4}}-\frac{4}{3}=\frac{2a^2}{2a^2}-\frac{4}{3}=\frac{2}{3} \] Therefore, the limit we are looking for is \(\boxed{\frac{2}{3}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
When we talk about indeterminate forms, we refer to expressions that do not immediately reveal a definite limit when we substitute the limiting value of the variable. A typical example is the expression \( \frac{0}{0} \), which occurs frequently in calculus when evaluating limits. Indeterminate forms can take several shapes such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) and can't be solved by direct substitution. The solution provided showcased one such indeterminate form when evaluating the limit at \( x = a \).

To resolve such cases, a range of techniques, including factoring, conjugate multiplication, or using the special L'Hôpital's Rule, can be applied. L'Hôpital's Rule is particularly handy when the conventional techniques are cumbersome or inapplicable, as it allows the simplification of the limit by differentiating the numerator and denominator separately.
Limits and Continuity
In understanding limits and continuity, we assume that if we can make the value of a function arbitrarily close to a certain value by choosing values close to a given point, then we say the limit of the function as it approaches that point equals that value. This concept is vital for defining continuity, where we expect a function to have no sudden jumps or breaks. In the context of the exercise, the function's limit is expected as \( x \to a \), which means we want to find the behavior of the function as \( x \) becomes very close to \( a \).

Ensuring this understanding, when we see the format \( \frac{0}{0} \)—an indication of an indeterminate form—we use L'Hôpital's Rule to find a limit that can reveal the true behavior of the function near that point. This process directly relates to the function's continuity at that point and its vicinity.
Differentiation
The process of differentiation is the keystone of calculus, which focuses on finding the rate at which things change. It provides us with the derivative of a function, which in turn, tells us how a function's value will change as its input changes. In the exercise, differentiation is the critical step in applying L'Hôpital's Rule. It demands a robust understanding of derivative rules, such as the product rule, quotient rule, and chain rule.

The solution applies these rules to both the numerator and denominator, leading to a new expression that can be more easily evaluated as \( x \to a \). The derivatives essentially simplify the limit problem to one where the limit can be determined without the indeterminate form \( \frac{0}{0} \), revealing the underlying behavior of the function around the point of interest and resolving the indeterminate form.

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