Problem 5 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 0} \frac{\sqrt{9+x}-3}{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Indeterminate Form

First, we need to substitute the value of the limit into the expression to see the form. Substituting \( x = 0 \), we get \( \frac{\sqrt{9+0}-3}{0} = \frac{3-3}{0} = \frac{0}{0} \), which is an indeterminate form.

Step 2: Apply the Conjugate Method

To resolve the indeterminate form, multiply the numerator and denominator by the conjugate of the numerator: \( \frac{\sqrt{9+x}-3}{x} \times \frac{\sqrt{9+x}+3}{\sqrt{9+x}+3} \).

Step 3: Simplify the Expression

Multiply the numerators and expand: \[ (\sqrt{9+x}-3)(\sqrt{9+x}+3) = (9+x) - 3^2 = 9 + x - 9 = x \] Now, the expression becomes \( \frac{x}{x(\sqrt{9+x}+3)} \).

Step 4: Cancel the Common Factor

In the expression \( \frac{x}{x(\sqrt{9+x}+3)} \), cancel the \( x \) in the numerator and the denominator, resulting in \( \frac{1}{\sqrt{9+x}+3} \).

Step 5: Evaluate the Limit

Now substitute \( x = 0 \) back into the simplified expression: \[ \lim_{x \to 0} \frac{1}{\sqrt{9+0}+3} = \frac{1}{3+3} = \frac{1}{6}. \]

Key Concepts

Indeterminate Forms

When dealing with limits, particularly those that involve division, you often encounter troublesome expressions known as "indeterminate forms." One common type is \( \frac{0}{0} \). This indicates that both the numerator and the denominator approach zero as the variable approaches a specific value. However, it doesn't tell us what the overall fraction approaches. This is problematic because direct substitution results in an ambiguous or undefined expression.
Understanding and identifying indeterminate forms is crucial because it tells us that the limit needs further analysis or manipulation to be evaluated. Techniques like algebraic manipulation or applying calculus rules (e.g., L'Hôpital's Rule) are employed to resolve the ambiguity and find the actual limit.

Conjugate Method

The conjugate method is a powerful algebraic technique used to simplify expressions, particularly those involving square roots. The "conjugate" of an expression like \( \sqrt{a} - b \) is \( \sqrt{a} + b \). This method involves multiplying the troublesome expression by a form of 1, often the numerator or denominator's conjugate over itself, such as \( \frac{\sqrt{a}+b}{\sqrt{a}+b} \).
Using the conjugate helps in rationalizing the denominator or numerator. It works by creating a difference of squares, which can likely be simplified further. In our problem, multiplying by the conjugate eliminated the square root, turning it into a form we could work with algebraically to simplify the limit problem.

Simplifying Expressions

Simplifying expressions is a key part of solving limit problems, especially when indeterminate forms are involved. It allows us to rearrange and reduce expressions to more manageable terms. Once the algebraic identities are applied, expressions often reduce to a simpler form.

• Identify common algebraic identities.
• Combine like terms to reduce complexity.
• Remove unnecessary parts, like zero terms.
• Factor expressions when possible.

In our task, simplifying was crucial after applying the conjugate method. Multiplying and simplifying the numerator resulted in cancellation, revealing the true nature of the limit. This step allows us to cancel terms and reveals the limit's behavior more straightforwardly.

Canceling Common Factors

Canceling common factors is an efficient way to simplify expressions further after applying methods like conjugation. It involves recognizing terms in the numerator and the denominator that are the same and removing them to simplify the fraction. This is possible when the factor is not zero because dividing by zero is undefined.
In our example, after simplifying the expression using the conjugate method, \( x \) appeared in both the numerator and denominator. By canceling this factor, we reduced the expression to a simpler form without changing the limit's value. This simplification is essential as it resolves the indeterminate form and leads to a piece that is easier to evaluate.