Evaluating the Limit as x Approaches 9: (√x - 3) / (x - 9)

The problem (√x - 3) / (x - 9) as x approaches 9 is a classic example of an indeterminate form, often represented in the form 0/0. This article explores the step-by-step method to solve this problem using both L'H?pital's Rule and algebraic manipulation, ensuring a deep understanding of the underlying concepts.

Understanding the Indeterminate Form

When we substitute x 9 directly into the expression (√x - 3) / (x - 9), both the numerator and the denominator evaluate to 0, leading to an indeterminate form 0/0. This requires a more sophisticated approach to determine the limit accurately.

Solving Using L'H?pital's Rule

L'H?pital's Rule is a powerful tool used to evaluate limits of indeterminate forms. According to this rule, if the limit of the ratio of the derivatives of the numerator and the denominator exists, then this limit is equal to the limit of the original expression.

lim_{x→9} (√x - 3) / (x - 9) lim_{x→9} (1 / (2√x)) 1 / 6

First, let's apply L'H?pital's Rule step-by-step:

Identify the derivative of the numerator √x - 3. The derivative is 1 / (2√x). Identify the derivative of the denominator x - 9. The derivative is 1. Apply L'H?pital's Rule: lim_{x→9} (1 / (2√x)). Evaluate the limit:

lim_{x→9} (1 / (2√x)) 1 / (2√9) 1 / 6

This result, 1 / 6, is the limit of the given expression.

Algebraic Manipulation Method

Another method involves algebraic manipulation to simplify the expression, making the limit easier to evaluate.

frac{√x - 3}{x - 9} frac{√x - 3}{(√x 3)(√x - 3)}

This step recognizes that the denominator can be factored using the difference of squares. Simplifying further:

frac{√x - 3}{(√x 3)(√x - 3)} frac{1}{√x 3}

Now, evaluate the limit as x → 9:

lim_{x→9} frac{1}{√x 3} frac{1}{√9 3} frac{1}{6}

This second method also confirms the limit is 1 / 6.

Conclusion

Both methods, L'H?pital's Rule and algebraic manipulation, result in the same limit: 1 / 6. This demonstrates the versatility of mathematical tools in solving complex limit problems. Understanding these methods not only provides a clear solution but also reinforces the underlying principles of calculus.

Understanding how to tackle such problems is crucial for students and professionals in fields where calculus is essential. Whether you're dealing with engineering, physics, economics, or other STEM disciplines, mastering these skills is essential.