Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7-18, find the indicated limit. In most cases, it will be wise to do some algebra first (see Example 2). $$ \lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3} $$

Step-by-Step Solution

Answer

The limit is 6.

Step 1: Identify Indeterminate Form

First, substitute the value \( x = 3 \) into the expression \( \frac{x^{2} - 9}{x - 3} \). This gives us \( \frac{3^{2} - 9}{3 - 3} = \frac{0}{0} \), which is an indeterminate form. This indicates that further algebraic manipulation is necessary to evaluate the limit.

Step 2: Factor the Numerator

Notice that the numerator \( x^2 - 9 \) is a difference of squares. Use the identity \( a^2 - b^2 = (a-b)(a+b) \) to factor it as follows: \( x^2 - 9 = (x - 3)(x + 3) \).

Step 3: Simplify the Expression

Now, substitute the factored form back into the expression: \( \frac{(x - 3)(x + 3)}{x - 3} \). Since \( x eq 3 \), we can cancel out the \( x - 3 \) terms. This simplifies the expression to \( x + 3 \).

Step 4: Evaluate the Limit

Now substitute \( x = 3 \) into the simplified expression \( x + 3 \): \( 3 + 3 = 6 \). Thus, the limit is \( 6 \).

Key Concepts

Indeterminate Forms

Indeterminate forms occur when the direct substitution of a limit results in an expression like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are called "indeterminate" because they don't immediately indicate the actual value of a limit. Instead, they signal that more algebraic manipulation is needed.

To evaluate such limits, you often need to manipulate the expression algebraically to bypass the\( \frac{0}{0} \) form. This can involve factoring, rationalizing, or using trigonometric identities, among other techniques.

Understanding indeterminate forms is crucial for successfully calculating limits in calculus because they frequently appear in problems that require finding the limit as a variable approaches a particular value. Being aware of this concept helps you recognize when you need to perform additional steps rather than attempting to solve the indeterminate limit directly.

Difference of Squares

The difference of squares formula is a highly useful algebraic identity. It states that \( a^2 - b^2 = (a-b)(a+b) \). This identity is particularly helpful when dealing with polynomial expressions in limit problems, as it allows you to factor complex terms into simpler binomial expressions.

For example, in our original exercise, the expression \( x^2 - 9 \) can be factored using \( 3^2 \) as \( (x-3)(x+3) \).

Using this identity helps to simplify the problem significantly. Once factored, it becomes easier to cancel out terms that otherwise cause indeterminate forms, such as \( x-3 \) in this case. This step often resolves the indeterminate issue and makes it possible to evaluate the limit.

Mastering this technique can save a lot of time and simplify many different calculus problems effectively.

Limit Evaluation Steps

Evaluating limits can be straightforward with the right approach. Here's a series of steps to guide you through the process:

• **Step 1: Identify the Indeterminate Form** - Substitute the limit point into the expression. If you find an indeterminate form like \( \frac{0}{0} \), further action is needed.


• **Step 2: Factor the Expression** - Look for ways to simplify the expression. In the exercise, we used the difference of squares to factor \( x^2-9 \) into \( (x-3)(x+3) \).


• **Step 3: Simplify the Expression** - After factoring, cancel out any common terms that are causing the indeterminate form. This makes the limit more approachable. For our example, canceling \( x-3 \) led to a simpler expression.


• **Step 4: Evaluate the Simplified Expression** - Now, substitute the limit value again into this simplified form. Here, substituting \( x=3 \) into \( x+3 \) gives us a limit of \( 6 \).

Practicing these steps consistently can help you build a strong foundation for solving limits in calculus, making it easier to address this common type of problem.