Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{x^{3}-3 x^{2}+x}{x^{3}-2 x} $$

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{2}\).

Step 1: Plug in the limit value

First, try substituting the value of the limit, which is \(x = 0\), into the expression. This gives \(\frac{0^3 - 3(0)^2 + 0}{0^3 - 2(0)} = \frac{0}{0}\). Since \(\frac{0}{0}\) is an indeterminate form, we need another approach.

Step 2: Basic Simplification

Factor out the common terms in the numerator and denominator. In the numerator, factor out \(x\) to get \(x(x^2 - 3x + 1)\). In the denominator, factor out \(x\) to get \(x(x^2 - 2)\). The expression becomes \(\frac{x(x^2 - 3x + 1)}{x(x^2 - 2)}\).

Step 3: Cancel the common terms

Cancel the common \(x\) in both the numerator and the denominator. The expression simplifies to \(\frac{x^2 - 3x + 1}{x^2 - 2}\).

Step 4: Re-evaluate the limit

Now that we have simplified the expression, substitute \(x = 0\) again. The expression becomes \(\frac{0^2 - 3(0) + 1}{0^2 - 2} = \frac{1}{-2}\).

Step 5: Determine the limit

The simplified expression evaluates to \(-\frac{1}{2}\) when \(x\) approaches 0. Hence, the limit is \(-\frac{1}{2}\).

Key Concepts

Indeterminate Forms

When encountering calculus problems involving limits, you may sometimes come across expressions like \( \frac{0}{0} \). This result is referred to as an indeterminate form. Indeterminate forms occur when the direct substitution of the limit value does not produce a definite result. These forms can arise in limits involving rational functions, exponentials, and more.

The most common indeterminate forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), among others such as \( 0 \times \infty \), \( 1^\infty \), and so on. It's important to recognize these forms because they indicate that the limit cannot be directly evaluated by substitution. Instead, additional algebraic manipulation, such as factoring, rationalizing, or using L'Hôpital's Rule, may be required to resolve the situation and find the limit.

In the exercise provided, substituting \( x = 0 \) resulted in \( \frac{0}{0} \), prompting us to apply further algebraic simplifications to evaluate the limit correctly.

Rational Functions

A rational function is a ratio of two polynomials. These functions take the form \( \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials. In calculus, rational functions frequently appear in limit problems and may lead to indeterminate forms if both the numerator and the denominator evaluate to zero at the same point.

Important properties of rational functions include:

• The degree of the numerator and the denominator can impact the behavior of the function.
• If the degree of the numerator is less than that of the denominator, the limit at infinity is typically zero.
• Vertical asymptotes occur where the denominator equals zero, excluding points canceled out by the numerator.

To evaluate the limit of a rational function, finding common terms between the numerator and the denominator is often helpful. In our exercise, factoring allowed the cancellation of \( x \), resolving the indeterminate form that initially appeared at \( x = 0 \).

Factoring in Algebra

Factoring is a key algebraic tool that simplifies mathematical expressions and is especially useful in handling indeterminate forms. It involves expressing a polynomial as a product of its factors, which are simpler polynomials or numbers.

Here are some fundamental factoring techniques:

• Common Factor: Identify and extract the greatest common factor (GCF) shared by all terms.
• Difference of Squares: Use the identity \( a^2 - b^2 = (a - b)(a + b) \).
• Trinomials: Rewrite trinomials by grouping or using a factorization formula.

Factoring simplifies complex expressions by reducing them to a form where terms may be canceled or manipulated more easily during limit evaluations or other operations. In the given solution, factoring \( x \) from both the numerator and the denominator was crucial in simplifying the expression and solving the limit.