Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the limits. \(\lim _{x \rightarrow 3^{-}} \frac{x^{2}-x-6}{x-3}\)

Step-by-Step Solution

Answer

The limit is 5.

Step 1: Factor the Numerator

Identify if the numerator \(x^2 - x - 6\) can be factored. The expression can be factored as \((x-3)(x+2)\).

Step 2: Simplify the Expression

Once factored, the expression becomes \(\frac{(x-3)(x+2)}{x-3}\). You can simplify this by canceling out the \(x-3\) from the numerator and denominator, resulting in the expression \(x + 2\).

Step 3: Evaluate the Limit

Evaluate the limit of the simplified expression \(x + 2\) as \(x\) approaches 3 from the left. Substituting \(x = 3\) into \(x + 2\) gives \(3 + 2 = 5\).

Step 4: Consider One-Sided Limit

Since the expression was given specifically as \(\lim _{x \rightarrow 3^{-}}\), and there are no discontinuities in \(x + 2\), the limit approaching from the left is also 5.

Key Concepts

Factorization

Factorization is a crucial technique in calculus, especially when dealing with limits of rational expressions. In our example, we start with the expression \( x^2 - x - 6 \). The goal of factorization here is to express this quadratic polynomial as a product of two binomials. To factor \( x^2 - x - 6 \), we look for two numbers that multiply to \(-6\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)). These numbers are \(-3\) and \(+2\). Hence, we can write:

• \(x^2 - x - 6 = (x - 3)(x + 2)\)

This simplification is essential because it allows us to cancel out the \(x-3\) term in the numerator and the denominator in subsequent steps. This cancellation will help address potential discontinuities and undefined expressions, leading us to find the limit successfully.

One-Sided Limits

One-sided limits are a reliable tool when investigating the behavior of a function as the variable approaches a specific value from one side only, either from the left or the right. The notation \( \lim _{x \rightarrow 3^{-}} \) suggests that we are interested in how the function behaves as \(x\) approaches 3 from the left side (values slightly less than 3).
In our exercise, the one-sided limit indicates we focus on inputs slightly less than 3. Therefore, simplifying the rational expression helps us determine the function's behavior as \(x\) nears 3. When we cancel and simplify correctly, the function should be continuous at this point, which makes evaluating the one-sided limit straightforward.

• Remember that if a one-sided limit exists and equals a number \(L\), the ordinary limit also exists.
• If they differ, though, the overall limit at that point does not exist.

Continuity

Continuity is a fundamental idea in calculus, referring to when a function is unbroken or uninterrupted over its domain. To be continuous, a function must meet three criteria at a point \(x = c\):

• Firstly, the function \(f(x)\) must be defined at \(c\).
• Secondly, the limit \( \lim _{x \to c} f(x) \) must exist.
• Lastly, the value of the function at \(c\) must equal the limit, i.e., \( f(c) = \lim _{x \to c} f(x) \).

In the case of \( \lim _{x \rightarrow 3^{-}} (x + 2) = 5 \), this limit is continuous since \(x+2\) is a polynomial and polynomials are continuous everywhere in their domain. Thus, as \(x\) approaches 3, the function behaves smoothly: both the one-sided limit from the left and the value of the function at \(x = 3\) are equal, confirming continuity at that point.

Simplification in Calculus

Simplification in calculus is an essential strategy that allows us to handle complex expressions, especially when evaluating limits. In the example provided, after factoring the numerator \( (x-3)(x+2) \), we simplify by cancelling the term \(x-3\) with the denominator, a step that simplifies the expression into \(x + 2\).
Simplification can help avoid indeterminate forms like \( \frac{0}{0} \). By simplifying, we turn our focus on evaluating the meaningful behavior of the function as it approaches the point of interest.

• Reducing complexity through simplification makes it easier to assess whether discontinuities exist.
• It also turns an undefined form into a manageable expression that can be evaluated directly.

Thus, simplification is a pivotal skill that alleviates confusion and aids in finding limits smoothly and correctly.