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The problem is asking for the limit:

\[
\lim_{x \to 1} \frac{x^{11} - 1}{x^7 - 1}
\]

The hint provides the factorization of \(x^n - 1\):

\[
x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + \cdots + x + 1)
\]

### Step 1: Factor both the numerator and the denominator

- Numerator: \(x^{11} - 1 = (x - 1)(x^{10} + x^9 + \cdots + x + 1)\)
- Denominator: \(x^7 - 1 = (x - 1)(x^6 + x^5 + \cdots + x + 1)\)

### Step 2: Cancel the common factor \((x - 1)\)

After factoring, both the numerator and denominator have a common factor of \((x - 1)\). Thus, we can cancel out this common term, leaving us with:

\[
\frac{x^{10} + x^9 + \cdots + x + 1}{x^6 + x^5 + \cdots + x + 1}
\]

### Step 3: Evaluate the limit as \(x \to 1\)

Now substitute \(x = 1\) into the simplified expression:

- Numerator: \(1^{10} + 1^9 + \cdots + 1 = 11\)
- Denominator: \(1^6 + 1^5 + \cdots + 1 = 7\)

Thus, the limit becomes:

\[
\frac{11}{7}
\]

### Final Answer:
\[
\lim_{x \to 1} \frac{x^{11} - 1}{x^7 - 1} = \frac{11}{7}
\]

Would you like more details or have any questions?

Here are some related questions:
1. What happens if we change the exponents in this type of limit problem?
2. How do other factoring techniques work in similar limit problems?
3. How does the cancellation of terms in limits help avoid indeterminate forms?
4. Can L'Hôpital's Rule be used to solve this limit, and how does it compare?
5. How does this problem change if the limit was approaching a different value, say \(x \to 2\)?

**Tip**: Factoring common terms in limit problems often simplifies them and helps resolve indeterminate forms like \(\frac{0}{0}\).

lim_{x→1} (x^11 - 1) / (x^7 - 1) (Hint: Factor x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ⋯ + x + 1))

Solve the limit problem lim_{x→1} (x^11 - 1) / (x^7 - 1) using polynomial factorization. This solution breaks down the factorization of x^n - 1 and evaluates the limit by simplifying the expression. Ideal for advanced high school or undergraduate calculus students.

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