Consider the product $P(x) = (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$. Use repeated applications of the difference of squares pattern to express $P(x)$ in closed form, and evaluate the limit $\lim_{x \to 1} \frac{P(x)}{x

Consider the product $P(x) = (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$ . Use repeated applications of the difference of squares pattern to express $P(x)$ in closed form, and evaluate the limit $\lim_{x \to 1} \frac{P(x)}{x - 1}$ .

Final answer

$P(x) = x^{16} - 1; \quad \lim_{x \to 1} \frac{P(x)}{x - 1} = 16$

Steps

Apply the Difference of Squares Pattern

The difference of squares formula states that for any terms $a$ and $b$ : $(a - b)(a + b) = a^2 - b^2$

We multiply the factors of $P(x)$ from left to right in pairs, noticing how each resulting difference of squares pairs with the subsequent factor:

Multiply the first two factors: $(x - 1)(x + 1) = x^2 - 1$
Multiply this result by the next factor, $(x^2 + 1)$ : $(x^2 - 1)(x^2 + 1) = (x^2)^2 - 1^2 = x^4 - 1$
Multiply by $(x^4 + 1)$ : $(x^4 - 1)(x^4 + 1) = (x^4)^2 - 1^2 = x^8 - 1$
Multiply by the final factor, $(x^8 + 1)$ : $(x^8 - 1)(x^8 + 1) = (x^8)^2 - 1^2 = x^{16} - 1$

Thus, the closed form of the product is: $P(x) = x^{16} - 1$

Pitfall to Avoid: When squaring an exponential term such as $(x^4)^2$ , multiply the exponents to get $x^{4 \cdot 2} = x^8$ , rather than squaring the exponent to get $x^{16}$ . The powers double at each stage: $1 \to 2 \to 4 \to 8 \to 16$ .

Evaluate the Limit

We wish to find: $\lim_{x \to 1} \frac{P(x)}{x - 1}$

Substitute the factored expression of $P(x)$ into the quotient: $\frac{P(x)}{x - 1} = \frac{(x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)}{x - 1}$

For $x \neq 1$ , cancel the common factor $(x - 1)$ : $\frac{P(x)}{x - 1} = (x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$

Now evaluate the limit by direct substitution as $x \to 1$ : $\lim_{x \to 1} \frac{P(x)}{x - 1} = (1 + 1)(1^2 + 1)(1^4 + 1)(1^8 + 1) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$

(Alternative perspective: The limit $\lim_{x \to 1} \frac{x^{16} - 1}{x - 1}$ is the definition of the derivative of $f(x) = x^{16}$ at $x = 1$ , giving $f'(1) = 16(1)^{15} = 16$ .)

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student adds exponents instead of multiplying powers when squaring terms in the difference of squares pattern, leading to an incorrect power of 10 and limit of 10.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
(x - 1)(x + 1) = x^2 - 1 $\implies$ (x^2 - 1)(x^2 + 1) = x^4 - 1
Step 2
(x^4 - 1)(x^4 + 1) = x^6 - 1
Correct
(x^4 - 1)(x^4 + 1) = (x^4)^2 - 1 = x^8 - 1
When applying $(a - b)(a + b) = a^2 - b^2$ with $a = x^4$ , the first term is $(x^4)^2 = x^{4 \cdot 2} = x^8$ , not $x^6$ . Powers must be multiplied by 2 rather than increased by 2.
Step 3
(x^6 - 1)(x^8 + 1) = x^{10} - 1 $\implies$ P(x) = x^{10} - 1
Correct
(x^8 - 1)(x^8 + 1) = (x^8)^2 - 1 = x^{16} - 1 $\implies$ P(x) = x^{16} - 1
Because of the previous exponent error, the powers no longer matched, leading to an incorrect closed-form polynomial.
Step 4
$\lim_{x $\to$ 1}$ \frac{x^{10} - 1}{(x - 1)} = 10(1)^9 = 10_
Correct
$\lim_{x $\to$ 1}$ \frac{x^{16} - 1}{x - 1} = 16(1)^{15} = 16_
Your limit evaluation method using the derivative of $x^n$ at $x = 1$ was mathematically sound, but it yielded 10 instead of 16 due to the incorrect exponent in $P(x)$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You recognized that $(x - 1)(x + 1)(x^2 + 1) \dots$ cascades via repeated difference of squares.

You correctly used the derivative definition of the limit of $\frac{x^n - 1}{x - 1}$ as $x \to 1$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Exponent rule error in power of a power

You wrote $(x^4)^2 = x^6$ , adding 2 to the exponent instead of multiplying $4 \times 2 = 8$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the power-of-a-power rule $(x^m)^n = x^{m \cdot n}$ with adding exponents or assuming an arithmetic progression of exponents ( $2, 4, 6, 8, 10$ ) instead of a geometric doubling ( $1, 2, 4, 8, 16$ ).

REASONING PATH

Your ACEpath teacher’s suggested approach

Power of a Power Rule: When raising a power to another power, multiply exponents: $(x^a)^b = x^{a \cdot b}$ .
Always compute $(x^4)^2 = x^{4 \cdot 2} = x^8$ , doubling the exponent at each difference-of-squares step.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the difference of squares: $(x^4 - 1)(x^4 + 1) = (x^4)^2 - 1 = x^8 - 1$ .
Multiply by the remaining factor: $(x^8 - 1)(x^8 + 1) = (x^8)^2 - 1 = x^{16} - 1$ , so $P(x) = x^{16} - 1$ .
Evaluate the limit: $\lim_{x \to 1} \frac{x^{16} - 1}{x - 1} = \left. \frac{d}{dx}(x^{16}) \right|_{x=1} = 16(1)^{15} = 16$ .

Final answer

P(x) = x^{16} - 1; $\lim_{x $\to$ 1}$ \frac{P(x)}{x - 1} = 16_

NEXT STEPS

Final Takeaways from Your Teacher

Review the power-of-a-power law: $(x^m)^n = x^{m \cdot n}$ .
Remember that in repeated difference of squares, exponents double at every stage: $1 \to 2 \to 4 \to 8 \to 16$ .

OverviewACEpath Teacher's Diagnosis

02 The student attempts direct substitution without canceling the indeterminate zero factor, incorrectly concluding the limit is undefined or 0/0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
(x - 1)(x + 1) = x^2 - 1 $\implies$ (x^2 - 1)(x^2 + 1) = x^4 - 1 $\implies$ (x^4 - 1)(x^4 + 1) = x^8 - 1 $\implies$ (x^8 - 1)(x^8 + 1) = x^{16} - 1 $\implies$ P(x) = x^{16} - 1
Step 2
$\lim_{x $\to$ 1}$ \frac{x^{16} - 1}{x - 1} = \frac{1^{16} - 1}{1 - 1} = \frac{0}{0}\text{, which does not exist}_
Correct
$\lim_{x $\to$ 1} $\frac{P(x)}{x - 1} = \$ \lim_{x \to 1} $(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1) = 2$ \cdot $2$ \cdot $2$ \cdot$ 2 = 16_
Direct substitution yielded the indeterminate form $\frac{0}{0}$ , which indicates that a common factor of $(x - 1)$ must be canceled before evaluating the limit.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly paired consecutive terms using $(a - b)(a + b) = a^2 - b^2$ .

You accurately simplified the full product to $P(x) = x^{16} - 1$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Indeterminate form misinterpretation

You treated the indeterminate form $\frac{0}{0}$ as a final answer meaning 'does not exist'.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing an undefined direct substitution value with an indeterminate limit signal. When direct substitution produces $\frac{0}{0}$ , the limit is not necessarily non-existent; rather, the factor $(x - a)$ causing the zero must be canceled from both numerator and denominator before re-evaluating.

REASONING PATH

Your ACEpath teacher’s suggested approach

Recognize Indeterminate Forms: Obtaining $\frac{0}{0}$ upon direct substitution means the result is indeterminate, not undefined.
Do not stop at $\frac{0}{0}$ ; look to factor and cancel the shared $(x - c)$ factor.
Leverage Existing Factorization: If the original function is already given in factored form, use that factored expression to divide out the denominator factor directly.
Substitute $P(x) = (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$ into $\frac{P(x)}{x - 1}$ , cancel $(x - 1)$ , and evaluate the remaining factors at $x = 1$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Write $\frac{P(x)}{x - 1}$ using the factored form of $P(x)$ : $\frac{(x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)}{x - 1}$ .
Cancel the common non-zero factor $(x - 1)$ for $x \neq 1$ : $(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$ .
Substitute $x = 1$ : $(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2^4 = 16$ .

Final answer

P(x) = x^{16} - 1; $\lim_{x $\to$ 1}$ \frac{P(x)}{x - 1} = 16_

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between undefined values and indeterminate limit forms such as $\frac{0}{0}$ .
Practice limit problems where dividing out common factors removes the zero in the denominator.
Notice that limits of the form $\lim_{x \to 1} \frac{x^n - 1}{x - 1}$ also represent the derivative of $f(x) = x^n$ at $x = 1$ , giving $n(1)^{n-1} = n$ .

OverviewACEpath Teacher's Diagnosis

03 The student cancels (x - 1) but adds the evaluated factors instead of multiplying them, obtaining 8 instead of 16.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$(x - 1)(x + 1) = x^2 - 1 \implies (x^2 - 1)(x^2 + 1) = x^4 - 1 \implies (x^4 - 1)(x^4 + 1) = x^8 - 1 \implies (x^8 - 1)(x^8 + 1) = x^{16} - 1 \implies P(x) = x^{16} - 1$
Step 2
$\lim_{x \to 1} \frac{P(x)}{x - 1} = \lim_{x \to 1} (x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$
Step 3
$= (1 + 1) + (1 + 1) + (1 + 1) + (1 + 1) \implies = 2 + 2 + 2 + 2 = 8$
Correct
$= (1 + 1)(1^2 + 1)(1^4 + 1)(1^8 + 1) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$
The canceled expression is a product of four factors, but you placed plus signs between the evaluated factors instead of multiplying them.

WHAT YOU DID WELL

Compliments from Your Teacher

You repeatedly applied the difference of squares identity $(a - b)(a + b) = a^2 - b^2$ to arrive at $P(x) = x^{16} - 1$ .

You correctly recognized that dividing by $(x - 1)$ cancels the first linear factor.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Operation Switch: Addition instead of Multiplication

You accidentally added the evaluated parenthetical terms $(2 + 2 + 2 + 2)$ rather than continuing the product $(2 \cdot 2 \cdot 2 \cdot 2)$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing multiplicative factors with additive terms during the direct substitution step of limit evaluation.

REASONING PATH

Your ACEpath teacher’s suggested approach

Preserve the Main Operation: If terms are grouped inside parentheses without an operator between them, they are multiplied: $(A)(B)(C) = A \cdot B \cdot C$ .
Always copy the implied multiplication symbols explicitely before substituting values to prevent turning products into sums.

CORRECTED PATH

ACEpath Teacher's Demonstration

Substitute $x = 1$ into each factor of the product:
$\lim_{x \to 1} (x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1) = (1 + 1)(1^2 + 1)(1^4 + 1)(1^8 + 1)$
Evaluate each numerical factor: $(2)(2)(2)(2)$
Multiply the factors: $2 \cdot 2 \cdot 2 \cdot 2 = 16$

Final answer

P(x) = x^{16} - 1; $\lim_{x $\to$ 1}$ \frac{P(x)}{x - 1} = 16_

NEXT STEPS

Final Takeaways from Your Teacher

Double-check your operator symbols when transitioning from factored algebraic expressions to numerical evaluation.
Practice using the derivative shortcut $\lim_{x \to 1} \frac{x^n - 1}{x - 1} = n(1)^{n-1} = n$ to verify your polynomial limit results.

Lock it in

Question

What is the result of simplifying $((x^a)^2)$ using exponent rules?

Answer

$x^{2a}$ . When raising a power to a power, you multiply the exponents: $(x^a)^b = x^{a \cdot b}$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student adds exponents instead of multiplying powers when squaring terms in the difference of squares pattern, leading to an incorrect power of 10 and limit of 10.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student attempts direct substitution without canceling the indeterminate zero factor, incorrectly concluding the limit is undefined or 0/0.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student cancels (x - 1) but adds the evaluated factors instead of multiplying them, obtaining 8 instead of 16.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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