Problem 31 - Precalculus

Question

Evaluate the following limits by rewriting the given expression as needed. $$\lim _{x \rightarrow 3} \frac{2 x^{2}-3 x-9}{x-3}$$

Step-by-Step Solution

Answer

The limit is 9.

Step 1: Identify the Problem Type

We are asked to evaluate the limit of a rational function as \(x\) approaches 3. Direct substitution appears to lead to an indeterminate form \(\frac{0}{0}\), so we need to rewrite the expression to evaluate the limit.

Step 2: Factor the Numerator

We begin by factoring the quadratic polynomial in the numerator, \(2x^2 - 3x - 9\). We need to find two numbers that multiply to \(-18\) (\(2\times -9\)) and add to \(-3\). These numbers are \(3\) and \(-6\). Factoring gives us: \(2x^2 - 3x - 9 = (2x + 3)(x - 3)\).

Step 3: Simplify the Expression

Now that the numerator is factored into \((2x + 3)(x - 3)\), we can simplify the expression by canceling out the \((x - 3)\) term from the numerator and denominator, assuming \(x eq 3\). This simplifies to \(\frac{2x + 3}{1} = 2x + 3\).

Step 4: Evaluate the Limit

With the simplified expression \(2x + 3\), we can now evaluate the limit as \(x\) approaches 3. Substituting \(x = 3\) into \(2x + 3\) gives us \(2(3) + 3 = 6 + 3 = 9\).

Key Concepts

Rational Function

A rational function is a fraction where both the numerator and the denominator are polynomials. In this case, we have the function \( \frac{2x^2 - 3x - 9}{x - 3} \). An important feature of rational functions is that they can become undefined at certain points, called vertical asymptotes, when the denominator equals zero. Hence, it is critical to find ways to simplify these functions, especially when evaluating limits.
A rational function can provide some interesting behavior in its graph as \(x\) approaches certain values. Learning how to manage these functions through techniques like factoring and simplifying is an essential skill for anyone studying calculus.

Factoring

Factoring is a powerful algebraic tool that simplifies polynomials. It involves expressing a polynomial as a product of simpler polynomials. In our exercise, we had a quadratic polynomial in the numerator, \(2x^2 - 3x - 9\). The challenge was to express this polynomial in a factored form.
To factor \(2x^2 - 3x - 9\), we found two numbers whose product is \(-18\) (from \(2\times -9\)) and whose sum is \(-3\). These numbers are 3 and \(-6\).

• Step 1: Rewrite \(2x^2 - 3x - 9\) as \((2x + 3)(x - 3)\).
• Step 2: Confirm by expanding to get back the original polynomial.

Through factoring, we simplify complex expressions and allow for easier manipulation, especially in limit evaluation.

Indeterminate Form

When trying to evaluate a limit directly, you might encounter what is called an "indeterminate form," such as \(\frac{0}{0}\). This happened when we substituted \(x = 3\) directly into \(\frac{2x^2 - 3x - 9}{x - 3}\). Instead of a clear result, it led to \(\frac{0}{0}\), which signifies that more algebraic manipulation is needed.
Indeterminate forms suggest that the function can be restructured—often by factoring or simplifying—to resolve the limit. They are a signal for us to look deeper into the expression to find a viable method to solve the problem correctly without hitting infinity or undefined values.

Simplification

Simplification is the process of rewriting an expression in a more manageable form. Once we have factored \(2x^2 - 3x - 9\) into \((2x + 3)(x - 3)\), we simplified the original rational function. Simplification involved canceling out the common term \((x - 3)\) in the numerator and denominator. This was possible under the assumption that \(xeq 3\) to avoid division by zero.

• Step 1: Identify the common term and cancel it, simplifying to \(2x + 3\).
• Step 2: Evaluate the limit using this simpler expression.

After simplification, substituting \(x = 3\) into \(2x + 3\) becomes straightforward, allowing for accurate evaluation of the limit, which equates to 9. Simplification transforms complex problems into solvable ones, making it a crucial technique in calculus.