Problem 100 - Calculus: Early Transcendentals

Question

Evaluate \(\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}\).

Step-by-Step Solution

Answer

Answer: \(\frac{1}{4}\)

Step 1: Write given function

The given function is \(\frac{\sqrt[4]{x}-2}{x-16}\) and we want to find its limit as \(x \rightarrow 16\).

Step 2: Multiply by the conjugate

In order to eliminate the fourth root in the numerator, we multiply both the numerator and the denominator by the conjugate of the numerator: \((\sqrt[4]{x}+2)\). \((\frac{\sqrt[4]{x}-2}{x-16}) \cdot (\frac{\sqrt[4]{x}+2}{\sqrt[4]{x}+2}) = \frac{(\sqrt[4]{x}-2)(\sqrt[4]{x}+2)}{(x-16)(\sqrt[4]{x}+2)}\)

Step 3: Use the difference of squares formula

Apply the difference of squares formula to the numerator and simplify. $(\sqrt[4]{x}-2)(\sqrt[4]{x}+2) = (\sqrt[4]{x})^2 - 2^2 = x - 4^2 \\ \frac{x-16}{(x-16)(\sqrt[4]{x}+2)}$

Step 4: Cancel common terms

The term \((x-16)\) can be cancelled from both numerator and denominator. \(\frac{x-16}{(x-16)(\sqrt[4]{x}+2)} = \frac{1}{\sqrt[4]{x}+2}\)

Step 5: Evaluate the limit

Now, evaluate the limit by plugging in the value \(x=16\). \(\lim_{x \rightarrow 16} \frac{1}{\sqrt[4]{x}+2} = \frac{1}{\sqrt[4]{16}+2} = \frac{1}{2+2} = \frac{1}{4}\) So, the limit of the given function as \(x \rightarrow 16\) is \(\boxed{\frac{1}{4}}\).

Key Concepts

Conjugate Method

When faced with complex expressions involving radicals during limit evaluations, the conjugate method is a valuable strategy. It involves multiplying the numerator and the denominator of a fractional expression by the conjugate of the numerator. This helps in rationalizing the numerator, effectively simplifying the expression.
Imagine you have an expression \(rac{\sqrt[4]{x} - 2}{x - 16}\). The conjugate of \(\sqrt[4]{x} - 2\) is \(\sqrt[4]{x} + 2\). By multiplying the entire expression by \(\frac{\sqrt[4]{x} + 2}{\sqrt[4]{x} + 2}\), you use the principle of multiplying by 1 to simplify without changing the value.
The result will be:

• The numerator transforms to a simpler form when using the difference of squares.
• The expression becomes more manageable, allowing further simplification.

Breaking down radicals in limits becomes easier and allows you to eliminate indeterminate forms.

Difference of Squares

The difference of squares is a simple yet powerful algebraic identity, which states that for any two numbers \((a - b)(a + b) = a^2 - b^2\). It is particularly useful in simplifying expressions during limit problems by removing the complexity of a square root or higher-degree roots.
In the provided exercise, you encounter the expression \(\sqrt[4]{x} - 2\) and its conjugate \(\sqrt[4]{x} + 2\). Using the difference of squares, this expands and simplifies to form: \((\sqrt[4]{x})^2 - 2^2 = x - 16\). This key simplification allows you to cancel terms effortlessly.
With this principle:

• You reduce the complexity of the expression.
• Enable the cancellation of common terms in the numerator and denominator.
• Simplify an expression to reach its limit effectively.

The difference of squares acts as a bridge in transforming complex analytical problems into simpler algebraic ones, crucially aiding the evaluation of limits.

Evaluating Limits

Evaluating limits is a foundational concept in calculus that helps in understanding the behavior of functions as they approach specific points. The goal is to find what value a function approaches as the input "tends" towards a certain number.
In our example, the limit we try to find is \(\lim_{x \rightarrow 16} \frac{\sqrt[4]{x} - 2}{x - 16}\). To resolve the indeterminate form \(\frac{0}{0}\), we use techniques like factoring, multiplying by conjugates, and simplifying expressions using algebraic identities.
The process involves:

• Transformations to eliminate the indeterminate form.
• Ultimately substituting the value of interest into the simplified expression.
• Arriving at a conclusive, finite limit value.

By substituting \(x = 16\) into the simplified expression \(\frac{1}{\sqrt[4]{x} + 2}\), you obtain the final limit as \frac{1}{4}\. Thus, evaluating limits not only provides insights into function behavior near tricky points but also guides toward a deeper understanding of continuity and infinite processes.