Question

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

Short answer

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

Step-by-step solution

Write given function

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

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: \((\sqrt[4]{x}+2)\). \((\frac{\sqrt[4]{x}-2}{x-16}) \cdot (\frac{\sqrt[4]{x}+2}{\sqrt[4]{x}+2}) = \frac{(\sqrt[4]{x}-2)(\sqrt[4]{x}+2)}{(x-16)(\sqrt[4]{x}+2)}\)

Use the difference of squares formula

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

Cancel common terms

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

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.