Problem 27 - Brief Applied Calculus

Question

Find the following limits without using a graphing calculator or making tables. $$ \lim _{x \rightarrow-1} \frac{3 x^{3}-3 x^{2}-6 x}{x^{2}+x} $$

Step-by-Step Solution

Answer

The limit is \(-9\).

Step 1: Check for Form Type

First, substitute \(x = -1\) into the expression to check the form of the limit:\[ \frac{3(-1)^3 - 3(-1)^2 - 6(-1)}{(-1)^2 + (-1)} = \frac{-3 - 3 + 6}{1 - 1} = \frac{0}{0} \]Thus, the limit is an indeterminate form of \( \frac{0}{0} \). This indicates we need to simplify the expression to resolve the limit.

Step 2: Factor the Expression

Factor both the numerator and the denominator of the expression where possible:The numerator is \(3x^3 - 3x^2 - 6x\) which can be factored by taking \(3x\) common:\[ 3x(x^2 - x - 2) \]Further factor \(x^2 - x - 2\) as \((x-2)(x+1)\), giving:\[ 3x(x-2)(x+1) \]The denominator \(x^2 + x\) can be factored as \(x(x + 1)\).The entire fraction becomes:\[ \frac{3x(x-2)(x+1)}{x(x+1)} \]

Step 3: Cancel Common Factors

Cancel out the common terms \(x(x+1)\) from both numerator and denominator:\[ \frac{3x(x-2)(x+1)}{x(x+1)} = 3(x-2) \]

Step 4: Substitute the Limit Value

Substitute \(x = -1\) into the expression obtained after cancellation:\(3(x-2)\) becomes:\[3(-1-2) = 3(-3) = -9\]Therefore, the limit is \(-9\).

Key Concepts

Indeterminate Forms

When evaluating limits in calculus, you'll often encounter indeterminate forms. These are expressions where the limit cannot be directly determined because they end up in a form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Such forms require further manipulation to find the limit.
For instance, when you plug \( x = -1 \) into the original problem, \( \frac{3x^3 - 3x^2 - 6x}{x^2 + x} \), it simplifies to \( \frac{0}{0} \), an indeterminate form. This tells us that direct substitution is not enough to evaluate the limit as it does not provide a conclusive result.
In such cases, we need to simplify the expression by using algebraic techniques like factoring or rationalizing, which can help in simplifying the expression further.

Factoring Polynomials

Factoring polynomials is a crucial step when dealing with indeterminate forms in limits. The idea is to simplify the expression to eliminate indeterminate forms.

• The numerator \( 3x^3 - 3x^2 - 6x \) is factored by extracting the greatest common factor, which is \( 3x \).
  This gives: \[ 3x(x^2 - x - 2) \]
• The trinomial \( x^2 - x - 2 \) can be further factored into \((x-2)(x+1)\).
• The denominator \( x^2 + x \) is factored by taking out \( x \), resulting in \( x(x+1) \).

With these factored forms, the expression simplifies to \[ \frac{3x(x-2)(x+1)}{x(x+1)} \].
This helps us in the next steps to cancel out common terms and move toward solving the limit.

Limit Evaluation Steps

After simplifying polynomials, limit evaluation becomes straightforward. These are the typical steps you follow:

• Cancel out any common factors in the numerator and denominator. Here, \( x(x+1) \) is canceled from both, reducing the expression to \( 3(x-2) \).
• Once the expression is simplified, substitute the limit value into the reduced form. For this exercise, substitute \( x = -1 \) into \( 3(x-2) \).

Performing the substitution, we get:
\[ 3(-1-2) = 3(-3) = -9 \]
And thus, the limit is confirmed to be \(-9\), resolving the indeterminate form and completing the evaluation.