Problem 65 - Precalculus

Question

Given \(f(x)=\frac{2 x}{x^{3}+216},\) find $$\lim _{x \rightarrow-6} f(x)$$

Step-by-Step Solution

Answer

\(\lim_{x \to -6} f(x) = \frac{1}{54}\)

Step 1: Identify the Limit Expression

The problem requires us to find the limit of the function \( f(x) = \frac{2x}{x^3 + 216} \) as \( x \to -6 \). Thus, we have: \[ \lim_{x \to -6} \frac{2x}{x^3 + 216} \] .

Step 2: Factor the Denominator

Notice that the denominator is of the form \( x^3 + 216 \), which is a sum of cubes: \( x^3 + 6^3 \). It can be factored as: \[ x^3 + 6^3 = (x + 6)(x^2 - 6x + 36) \].

Step 3: Substitute the Limit Value

Substitute \( x = -6 \) into the denominator: \((x + 6) = 0\). This confirms the need to apply L'Hospital's Rule since the fraction becomes \( \frac{0}{0} \).

Step 4: Determine Applicability of L'Hospital's Rule

Since both the numerator \( 2x \) and the denominator \( x^3 + 216 \) both are zero at \( x = -6 \), L'Hospital's Rule is applicable.

Step 5: Apply L'Hospital's Rule

Differentiate the numerator and the denominator. The derivative of \( 2x \) is \( 2 \). The derivative of \( x^3 + 216 \) is \( 3x^2 \). Using L'Hospital's Rule, evaluate: \[ \lim_{x \to -6} \frac{2}{3x^2} \]

Step 6: Evaluate the Limit

Now simplify and evaluate the expression: \[ \lim_{x \to -6} \frac{2}{3(-6)^2} = \frac{2}{108} = \frac{1}{54} \].

Key Concepts

L'Hospital's Rule

L'Hospital's Rule is a very useful tool in calculus for finding limits of indeterminate forms, like "\( \frac{0}{0} \)" or "\( \frac{\infty}{\infty} \)." When you encounter such forms in a limit, it means that directly substituting the value into the function won't work because it doesn't give a clear answer. Instead, L'Hospital's Rule lets you differentiate the top and bottom of your fraction separately to try finding the limit again.
To use L'Hospital's Rule:

• Ensure that when you substitute the value, both the numerator and denominator give zero (or both give infinity). This confirms you have an indeterminate form.
• Differentiate the numerator and the denominator separately.
• Evaluate the limit again using these derivatives.

This method keeps going if needed, sometimes requiring multiple applications if the first derivative still leads to an indeterminate form. The solution of the provided exercise using L'Hospital's Rule shows how this approach helps in finding clear answers even when things initially seem undefined.

Sum of Cubes

The expression \( x^3 + 216 \) in the exercise showcases the concept of a sum of cubes. Recognizing patterns is important because they help simplify expressions.

• A sum of cubes like \( x^3 + 6^3 \) can be factored using the formula: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).
• In our case, \( x^3 + 6^3 = (x + 6)(x^2 - 6x + 36) \).
• Factoring helps identify and cancel common terms when evaluating limits, which is often essential for applying strategies like L'Hospital's Rule.

By converting it into a factored form, it became clear which terms to study when substituting \( x = -6 \). Observing that \( x + 6 = 0 \) when \( x = -6 \) indicates why the original limit was an indeterminate form, thus requiring additional steps to solve.

Derivatives

Derivatives are a fundamental concept in calculus and they offer the power to determine how a function behaves, primarily by studying how it changes. This means they tell us about the slopes of curves and rates of change. For the exercise here:

• The derivatives were taken for both the numerator and the denominator to resolve the indeterminate form identified earlier.
• Specifically, the derivative of the simple linear function \( 2x \) is a constant \( 2 \).
• The derivative of the polynomial \( x^3 + 216 \) is \( 3x^2 \), owing to applying basic differentiation rules.

By using these derivatives in L'Hospital's Rule, we transform the complex initial problem into something manageable and straightforward. Understanding derivatives not only helps solve this limit but also strengthens overall calculus problem-solving abilities. The entire process of differentiation aids immensely in finding limits, especially when direct evaluation leads to no conclusion.