Problem 368 - Calculus Volume 1

Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x+3}\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify Indeterminate Form

Before applying any method, determine if the expression forms an indeterminate type. Substitute \(x = 3\) into the function: \(\frac{3^2 - 9}{3 + 3} = \frac{0}{6} = 0\). Since the numerator becomes 0, it's not an indeterminate form and no further calculation is needed.

Step 2: Cancel Common Terms

Notice that the numerator \(x^2 - 9\) is a difference of squares, which can be factored as \((x - 3)(x + 3)\). The expression then becomes \(\frac{(x - 3)(x + 3)}{x + 3}\).

Step 3: Simplify the Expression

Since \(x + 3\) is in both the numerator and the denominator, cancel it out: \(\frac{(x - 3)(x + 3)}{x + 3} = x - 3\), for \(x eq -3\).

Step 4: Evaluate the Limit

Substitute \(x = 3\) into the simplified expression \(x - 3\): Substitute gives \(3 - 3 = 0\). Thus, the value of the limit is 0.

Key Concepts

Indeterminate Forms

Indeterminate forms appear in limit evaluation when both the numerator and the denominator of a fraction approach zero or infinity as the input approaches a specific value. This makes the limit undefined in its existing form, hence the term "indeterminate." Understanding indeterminate forms is essential when working with limits because they signal that the expression requires further simplification before applying mathematical techniques. Common indeterminate forms include:

• 0/0
• ∞/∞
• 0×∞
• ∞−∞
• 0⁰

In our exercise, we found that the initial substitution resulted in 0 in the numerator. However, because the denominator was a non-zero number, it did not result in an indeterminate form. Instead, it was a potentially solvable limit since the indeterminate form 0/0 did not arise.

Limit Evaluation

Evaluating limits often requires determining the behavior of a function as the input approaches a particular value. This is the core of calculus and crucial for understanding continuity and differentiability. There are multiple techniques for evaluating limits, including direct substitution, factorization, rationalization, and using L'Hôpital's Rule. For limits that result in indeterminate forms like 0/0, further techniques must be applied to resolve them.
In the original problem, direct substitution initially proved fruitless, but by factoring the polynomial in the numerator, and canceling common terms, it became possible to simplify the expression further. This allowed us to substitute the value accurately in the simplified expression to find that the limit was 0. Remember, not all limits require advanced techniques, but recognizing which approach to take is key.

Factoring Polynomials

Factoring polynomials is an algebraic technique to simplify expressions, especially useful in limit evaluation. When a polynomial can be factored into simpler expressions, it can reveal more about the function's behavior near a point of interest. For instance, in our problem, the polynomial \(x^2 - 9\) was factored into \((x - 3)(x + 3)\) using the rule for the difference of squares. Factoring made canceling common terms straightforward, simplifying the limit evaluation process. The general process for factoring a difference of squares is to recognize it in expressions of the form \(a^2 - b^2\) and factor it into \((a-b)(a+b)\). By mastering such algebraic techniques, solving limits becomes a more intuitive and efficient practice.