Problem 49 - Single Variable Calculus

Question

Find \(\lim _{x \rightarrow 2}\left|\frac{x^{2}+x-6}{x-2}\right|\)

Step-by-Step Solution

Answer

The limit of the function \(f(x) = \left|\frac{x^2 + x - 6}{x - 2}\right|\) as \(x\) approaches 2 is 5.

Step 1: Factor the numerator of the fraction

Factor the quadratic in the numerator: \(x^2 + x - 6\). To do this, we look for two numbers whose product is equal to -6 and whose sum is equal to 1, the coefficient of the linear term. These two numbers are 3 and -2. So, we can rewrite the quadratic as follows: \(x^2 + x - 6 = (x + 3)(x - 2)\)

Step 2: Simplify the function

Now, substitute the factored form of the numerator back into the function, \(f(x)\): \(f(x) = \left|\frac{(x + 3)(x - 2)}{x - 2}\right|\) Since the function is inside the absolute value, the function's output will always be non-negative. Notice that both the numerator and denominator contain the factor (x - 2). We can cancel this common factor out: \(f(x) = |x + 3|\)

Step 3: Find the limit as \(x\) approaches 2

Now that the function is simplified, we can find the limit as \(x\) approaches 2: \(\lim_{x \rightarrow 2} |x + 3|\) To evaluate this limit, we can simply substitute 2 into the function: \(|2 + 3| = |5| = 5\) Thus, the limit of the function as \(x\) approaches 2 is 5.

Key Concepts

Factoring Quadratics

When solving problems involving limits, such as finding \(\lim _{x \rightarrow 2}\left|\frac{x^{2}+x-6}{x-2}\right|\), factoring quadratics is often a crucial first step. Factoring involves rewriting a quadratic expression as a product of two binomials. For example, the quadratic expression \(x^2 + x - 6\) can be rewritten in a factored form.

To factor the quadratic, we need to identify two numbers that have a product equal to the constant term, -6, and a sum equal to the coefficient of the linear term, which is 1. This pair of numbers is 3 and -2.

• The product of 3 and -2 is \(-6\).
• Their sum is \(1\).

Therefore, the quadratic \(x^2 + x - 6\) can be factored as \((x + 3)(x - 2)\). This simplification helps us manipulate and simplify expressions, which is useful for solving limits.

Absolute Value Function

An absolute value function provides the distance of a number from zero on a number line, making any expression within the absolute value always non-negative. In our limit problem, we have an expression inside the absolute value: \(|f(x)| = \left| \frac{(x + 3)(x - 2)}{x - 2} \right|\).

When simplifying, we notice that both the numerator and the denominator have a common factor: \((x-2)\). Canceling these common factors leads to the expression \(|x + 3|\), which indicates that the function gives non-negative outputs depending directly on \(x+3\).

• The factorization and cancellation should be done cautiously to avoid any undefined terms.
• In absolute value, the expression’s sign is not influenced before simplifying.
• The essence of using absolute value here ensures manipulation doesn’t falsely change the outcome’s sign.

This function tells the actual distance of \((x+3)\) points from zero on the number line unaffected by negative signs, crucial for evaluating limits.

Simplifying Rational Functions

Simplifying rational functions is a powerful technique in calculus to make otherwise complicated expressions manageable. In our exercise, the rational function \(\frac{x^2 + x - 6}{x - 2}\) needed simplification.

By factoring the quadratic and canceling common terms, the expression simplifies to \(|x + 3|\). This new form is easier to evaluate as \(x\) approaches specific values.

• Rational functions involve both polynomial numerators and denominators.
• Simplification helps expose behaviors of functions near points of interest.
• We cautiously factor and cancel only matching terms in numerators and denominators.

The simplification makes calculating limits straightforward by reducing unnecessary complexity while maintaining correctness. It’s an indispensable step not only for finding limits but understanding function behavior in both calculus and algebra.