Problem 27 - Calculus and its applications

Question

The initial substitution of \(x=a\) yields the form \(0 / 0 .\) Look for ways to simplify the function algebraically, or use a table or graph to determine the limit. When necessary, state that the limit does not exist. $$ \lim _{x \rightarrow 25} \frac{\sqrt{x}-5}{x-25} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{10} \).

Step 1: Initial Substitution

First, try substituting the given value of \(x=25\) in the expression \( \frac{\sqrt{x}-5}{x-25} \).If you substitute directly:\( \frac{\sqrt{25}-5}{25-25} = \frac{0}{0} \),this gives the indeterminate form \( \frac{0}{0} \). Therefore, the limit needs to be evaluated by another method.

Step 2: Simplify the Function

To find the limit, we need to simplify the function by rationalizing the numerator.Multiply the numerator and the denominator by the conjugate of the numerator, \( \sqrt{x} + 5 \):\[ \frac{\sqrt{x}-5}{x-25} \times \frac{\sqrt{x}+5}{\sqrt{x}+5} = \frac{(\sqrt{x} - 5)(\sqrt{x} + 5)}{(x - 25)(\sqrt{x} + 5)} \]The numerator becomes:\( (\sqrt{x} - 5)(\sqrt{x} + 5) = x - 25 \).Now, the function simplifies to:\[ \frac{x - 25}{(x - 25)(\sqrt{x} + 5)} = \frac{1}{\sqrt{x} + 5} \] for \( x eq 25 \).

Step 3: Evaluate the Simplified Limit

Now that the function is simplified to \( \frac{1}{\sqrt{x} + 5} \), substitute \( x=25 \) into the simplified expression:\[ \frac{1}{\sqrt{25} + 5} = \frac{1}{5 + 5} = \frac{1}{10} \].Hence, the limit is \( \frac{1}{10} \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions where substituting a limit into a function results in an ambiguous value. One of the most common indeterminate forms is \( \frac{0}{0} \). This occurs when both the numerator and the denominator of a function approach zero, creating ambiguity about the limit's behavior.
Understanding indeterminate forms is crucial because they indicate that a direct substitution isn't enough to evaluate the limit. Instead, more sophisticated algebraic techniques or alternative methods, like L'Hôpital's Rule or simplification techniques, need to be employed.
Identifying an indeterminate form is the first step in determining whether a limit exists or if it requires further analysis. For example, in the exercise, substituting \(x = 25\) into the function \( \frac{\sqrt{x} - 5}{x - 25} \) initially results in \( \frac{0}{0} \), signaling that simplification or an alternative approach is necessary to find the limit.

Rationalizing Techniques

Rationalizing techniques involve converting a function into a form that avoids indeterminate results, simplifying the calculation of limits. This often includes multiplying by a conjugate, which is especially useful when dealing with square roots.

• Finding the Conjugate: The conjugate of \( \sqrt{x} - 5 \) is \( \sqrt{x} + 5 \). By multiplying both the numerator and the denominator by this conjugate, we leverage the difference of squares formula.
• Apply the Difference of Squares: This results in \( (\sqrt{x} - 5)(\sqrt{x} + 5) = x - 25 \) in the numerator, effectively eliminating the square root and leaving a factor of the original expression that can be canceled with the denominator.

Rationalizing transforms the original form \( \frac{0}{0} \) into a more manageable form that helps in evaluating the limit. In the exercise, this technique simplified the expression to \( \frac{1}{\sqrt{x} + 5} \), paving the way for easy limit computation.

Simplifying Expressions

Simplifying expressions in calculus is key to solving limits, especially when dealing with indeterminate forms. The heart of simplification lies in algebraic manipulation that transforms complex expressions into simpler ones. This often involves:

• Factoring: Break down complex terms into products of simpler factors.


• Canceling Terms: After factoring, any common terms in the numerator and denominator can usually be canceled.

In the exercise, simplifying the expression was achieved by rationalizing, which transformed the function from \( \frac{\sqrt{x} - 5}{x - 25} \) to \( \frac{1}{\sqrt{x} + 5} \). This form allowed direct substitution of \( x = 25 \) without issues, as the expression no longer resulted in an undefined form.Simplifying expressions using techniques like rationalization provides clarity and ease in the computation of limits and is an invaluable skill in calculus problem-solving.