Problem 23 - Calculus Early Transcendentals

Question

Evaluate the limit, if it exists. $$\lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{6} \).

Step 1: Direct Substitution

First, try directly substituting the value of 7 into the expression for \( x \). This means evaluating \( \frac{\sqrt{7+2}-3}{7-7} \). This results in \( \frac{\sqrt{9} - 3}{0} \), which simplifies to \( \frac{3 - 3}{0} = \frac{0}{0} \). Since this is an indeterminate form, we need to use another method to evaluate the limit.

Step 2: Rationalization

To resolve the indeterminate form, let's rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator. The conjugate is \( \sqrt{x+2}+3 \). The expression becomes: \[ \frac{\sqrt{x+2}-3}{x-7} \cdot \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3} = \frac{(\sqrt{x+2})^2 - 3^2}{(x-7)(\sqrt{x+2}+3)} \].

Step 3: Simplify the Expression

Simplify the expression obtained after rationalization: \[ \frac{x+2 - 9}{(x-7)(\sqrt{x+2}+3)} = \frac{x-7}{(x-7)(\sqrt{x+2}+3)} \].

Step 4: Cancel Common Factors

Notice that \( x-7 \) in the numerator and denominator are common factors. Cancel these out to get: \[ \frac{1}{\sqrt{x+2}+3} \].

Step 5: Re-evaluate the Limit

Now, substitute \( x = 7 \) into the simplified expression: \[ \frac{1}{\sqrt{7+2}+3} = \frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6} \].

Key Concepts

Indeterminate forms

When evaluating limits in calculus, you might come across expressions that result in an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not provide enough information to determine the limit directly. Instead of offering a clear result, they signal the need for further manipulation or alternative methods to resolve the limit.
Indeterminate forms occur when both the numerator and the denominator approach zero or infinity simultaneously as the variable tends to a particular value. This means simply substituting the value in doesn't give a meaningful answer.

• Examples of indeterminate forms include: \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), among others.
• They require techniques like factoring, rationalizing, or L'Hôpital's rule to resolve.

Understanding the nature of indeterminate forms is crucial for applying the correct method to resolve them and finding the actual limit.

Rationalization technique

Rationalization is a helpful technique used to eliminate indeterminate forms that involve square roots (or other roots). In the context of limits, the technique involves multiplying the numerator and the denominator by the conjugate of the irrational expression. This helps to simplify the irrational expression into a more manageable form.
The idea is that by multiplying by the conjugate, the roots get squared, simplifying the expression significantly. Conjugates are pairs like \( \sqrt{x+2} + 3 \) and \( \sqrt{x+2} - 3 \).

• For example, given \( \frac{\sqrt{x+2} - 3}{x-7} \), we multiply by \( \frac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3} \).
• This results in a difference of squares in the numerator: \( (\sqrt{x+2})^2 - 3^2 \), simplifying to simpler expressions.
• Rationalization transforms complicated or undefined forms into expressions we can evaluate comfortably.

Mastering rationalization enables effectively solving limits that initially seem complex due to root expressions.

Direct substitution method

The direct substitution method for finding limits is the simplest technique, often the first line of attack. It involves plugging in the value the variable is approaching directly into the given function and calculating the outcome.
This method is straightforward for continuous functions at the point of interest and gives immediate answers when applicable.

• For example, in \( \lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7} \), substituting \( x = 7 \) gives \( \frac{0}{0} \), which indicates an indeterminate form.
• When direct substitution leads to such forms, we resort to other techniques (like rationalization or factoring).
• If the simplification resolves the indeterminate nature, you can often return to a simple direct substitution.

Understanding when and how direct substitution works is essential for evaluating limits efficiently and knowing when you need to explore alternative methods.