Problem 47 - Calculus - AP Edition

Question

(a) Find the error in the following calculation: $$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^{3}-x^{2}+x-1}{x^{3}-x^{2}} &=\lim _{x \rightarrow 1} \frac{3 x^{2}-2 x+1}{3 x^{2}-2 x} \\ &=\lim _{x \rightarrow 1} \frac{6 x-2}{6 x-2}=1 \end{aligned}$$ (b) Find the correct limit.

Step-by-Step Solution

Answer

The error was in inappropriate application of derivative; the correct limit is 2.

Step 1: Evaluate the Original Expression

First, substitute the value 1 into the given expression: \[ \frac{x^3 - x^2 + x - 1}{x^3 - x^2} \] This reduces to \( \frac{1 - 1 + 1 - 1}{1 - 1} = \frac{0}{0} \), which is an indeterminate form, meaning we can't directly evaluate it.

Step 2: Incorrect Differentiation

Assess the step that led to: \[ \lim_{x \to 1} \frac{3x^2 - 2x + 1}{3x^2 - 2x} \] The student incorrectly applied the derivative in parts of the numerator and denominator separately without using an appropriate rule, such as L'Hôpital's Rule, which is applicable for \(\frac{0}{0}\) forms.

Step 3: Verify Using L'Hôpital's Rule

Since the original expression is indeterminate, apply L'Hôpital's Rule:Differentiate the numerator: \[ 3x^2 - 2x + 1 \rightarrow 3x^2 - 2x + 1 Differentiating gives: 6x - 2 \]Differentiate the denominator: \[ 3x^2 - 2x \rightarrow 6x - 2 \]Now the limit becomes:\[ \lim_{x \to 1} \frac{6x - 2}{6x - 2} = \lim_{x \to 1} 1 = 1 \]

Step 4: Correct the Limit

Factor both the numerator and denominator to simplify. The numerator \(x^3 - x^2 + x - 1\) can be rewritten as \((x-1)(x^2 + 1)\).The denominator \(x^3 - x^2\) is \(x^2(x-1)\).Cancel \((x-1)\) from top and bottom:\[ \lim_{x \to 1} \frac{x^2 + 1}{x^2} = \lim_{x \to 1} \left(1 + \frac{1}{x^2}\right) \]\Substitute \(x=1\) to get:\[ 1 + 1 = 2 \]

Key Concepts

Indeterminate Forms

An indeterminate form arises when an expression does not clearly define a specific numeric value through substitution. They often appear in calculus during limit calculations. In particular, the \( \frac{0}{0} \) form, as seen in our example, is one of the most common indeterminate forms encountered.
To address an indeterminate form, we typically need to manipulate the expression. This could involve algebraic simplification, factoring, or using calculus techniques like L'Hôpital's Rule.

• An indeterminate form signals that direct substitution does not yield a meaningful result.
• It's a cue to apply advanced techniques for limit evaluation.

Differentiation

Differentiation is a fundamental concept in calculus, referring to the process of finding the derivative of a function. The derivative measures how a function changes as its input changes, capturing the rate of change.
When dealing with limits of indeterminate forms, we may use differentiation to make further progress. However, in our problem, an incorrect attempt was initially made by differentiating separately parts of the expression.

• Correct application of differentiation can be guided by rules like the product rule, quotient rule, or chain rule.
• When L'Hôpital's Rule is applicable (such as in \( \frac{0}{0} \) forms), it allows us to differentiate the numerator and denominator to evaluate the limit.

Limit Calculation

Calculating limits is a crucial technique in understanding the behavior of functions as the input approaches a particular value. In the solved exercise, limits help us understand how the expression behaves near the point x = 1. The initial substitution led to the indeterminate form \( \frac{0}{0} \), but through L'Hôpital's Rule, a new limit was found using differentiation of the numerator and the denominator.
To find the correct limit, simplifying by factoring allowed for cancellation of terms that caused indeterminacy.

• Always check for ways to simplify expressions before attempting differentiation or applying L'Hôpital's Rule.
• Finding the limit of a function can reveal important characteristics about the function's behavior and continuity near specific points.