Question

Let \( f(x) = (x^5 - 1)(x^3 + 1) \), \( g(x) = (x^2 - 1)(x^2 - x + 1) \) and let \( h(x) \) be such that \( f(x) = g(x)h(x) \). Then \( \lim_{x \to 1} h(x) \) is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 3 \)
• D. \( 4 \)
• E. \( 5 \)

Hint:

When faced with \( (x^n - 1) / (x - 1) \), the result is always \( n \) as \( x \to 1 \). This follows from the sum of the geometric series \( 1 + x + \dots + x^{n-1} \).

Answer 1

Answer: (E)

Concept: We express \( h(x) = \frac{f(x)}{g(x)} \). To find the limit as \( x \to 1 \), we must simplify the expression to remove the indeterminate form \( 0/0 \) by factoring the polynomials.

Step 1: Factorize the components of \( f(x) \) and \( g(x) \).
Factorization identities:
• \( x^5 - 1 = (x-1)(x^4 + x^3 + x^2 + x + 1) \)
• \( x^3 + 1 = (x+1)(x^2 - x + 1) \)
• \( x^2 - 1 = (x-1)(x+1) \)

Step 2: Substitute these into the expression for \( h(x) \).
\[ h(x) = \frac{(x-1)(x^4 + x^3 + x^2 + x + 1) \cdot (x+1)(x^2 - x + 1)}{(x-1)(x+1) \cdot (x^2 - x + 1)} \]

Step 3: Cancel common factors and evaluate the limit.
The terms \( (x-1), (x+1), \) and \( (x^2 - x + 1) \) cancel out: \[ h(x) = x^4 + x^3 + x^2 + x + 1 \] \[ \lim_{x \to 1} h(x) = 1^4 + 1^3 + 1^2 + 1^1 + 1 = 5 \]