Problem 27 - Calculus: Early Transcendentals

Question

Evaluate the limit, if it exists. \( \displaystyle \lim_{x \to 16}\frac{4 - \sqrt{x}}{16x - x^2} \)

Step-by-Step Solution

Answer

The limit is \( \frac{1}{128} \).

Step 1: Analyze the given limit

We need to evaluate the limit \( \lim_{x \to 16}\frac{4 - \sqrt{x}}{16x - x^2} \). Substitute \( x = 16 \) directly to check if the expression is defined. By doing so, we get \( \frac{4 - \sqrt{16}}{16 \times 16 - 16^2} = \frac{0}{0} \), which is an indeterminate form. Therefore, further analysis is needed.

Step 2: Simplify the numerator

The numerator is \( 4 - \sqrt{x} \). To simplify, we can multiply and divide by the conjugate, which is \( 4 + \sqrt{x} \). This gives us:\[\frac{(4 - \sqrt{x})(4 + \sqrt{x})}{(16x - x^2)(4 + \sqrt{x})} \]Using the difference of squares, the numerator becomes \( (4)^2 - (\sqrt{x})^2 = 16 - x \).

Step 3: Simplify the denominator

The denominator \( 16x - x^2 \) can be factored. Notice that \( 16x - x^2 = x(16 - x) \). Thus, the expression becomes:\[\frac{16 - x}{x(16 - x)(4 + \sqrt{x})}\]

Step 4: Cancel common factors

Observe that \( 16 - x \) in the numerator and denominator can be canceled out, assuming \( x eq 16 \) during simplification:\[\frac{1}{x(4 + \sqrt{x})}\]

Step 5: Evaluate the simplified expression

Now substitute \( x = 16 \) into the simplified expression:\[\frac{1}{16(4 + \sqrt{16})} = \frac{1}{16(4 + 4)} = \frac{1}{16 \times 8}\]This evaluates to \( \frac{1}{128} \).

Step 6: Conclusion

The limit \( \lim_{x \to 16}\frac{4 - \sqrt{x}}{16x - x^2} \) is \( \frac{1}{128} \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms occur when directly substituting a limit into a function results in expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0 \cdot \infty \). These forms don't have a straightforward value, which is why they're "indeterminate." For the exercise, when we substitute \( x = 16 \) into the original limit expression \( \frac{4 - \sqrt{x}}{16x - x^2} \), we end up with \( \frac{0}{0} \), an indeterminate form.
This tells us that the given limit needs to be approached with a method that can resolve this ambiguity, like factoring or using the conjugate multiplication method. Recognizing these forms is crucial because it signals the need to simplify the expression further to find the actual limit.

Conjugate Multiplication

The conjugate multiplication method is often employed to simplify expressions involving square roots. The conjugate of an expression \( a - b \) is \( a + b \), and multiplying an expression by its conjugate utilizes the difference of squares identity \( a^2 - b^2 \).
In this exercise, the numerator \( 4 - \sqrt{x} \) was problematic. By multiplying by its conjugate \( 4 + \sqrt{x} \), you obtain \( (4 - \sqrt{x})(4 + \sqrt{x}) = 16 - x \). This process eliminates the square root and simplifies the evaluation. The conjugate multiplication converts complex expressions into a form easier to work with, especially when handling indeterminate forms.

Factoring

Factoring is another powerful technique in calculus for simplifying expressions, particularly polynomials in this context. The goal is to rewrite an expression as a product of simpler terms that are easier to manage. In the denominator of our exercise, the expression \( 16x - x^2 \) can be factored as \( x(16 - x) \).
By expressing the denominator this way, you identify that \( 16 - x \) appears as a factor in both the numerator and the denominator. This resemblance allows us to simplify by canceling out matching terms, assuming \( x eq 16 \). Factoring not only simplifies complex expressions but also uncovers cancelation opportunities, making limits more straightforward to evaluate.

Evaluating Limits

Evaluating limits is a fundamental concept in calculus, involving finding the value that a function approaches as the input approaches a certain point. Once an expression is simplified and indeterminate forms are resolved, evaluating the limit becomes more direct.
In the final step of our exercise, after simplifying \( \frac{4 - \sqrt{x}}{16x - x^2} \) into \( \frac{1}{x(4 + \sqrt{x})} \), substituting \( x = 16 \) gives us \( \frac{1}{128} \). This process shows how initially complex limits become straightforward as we manipulate and simplify the expressions. Mastering methods like substitution and simplifications ensures you're well-prepared to tackle a variety of limit problems in calculus.