The given problem is:

$\lim_{x \to a} \frac{x^n - a^n}{x^m - a^m}$

We will solve this limit using algebraic techniques.

Step-by-Step Solution

1. Factorization: Both the numerator and the denominator can be factored using the difference of powers formula. For $x^n - a^n$ and $x^m - a^m$, we use the following factorizations:

   $x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + \cdots + a^{n-1})$

   Similarly,

   $x^m - a^m = (x - a)(x^{m-1} + x^{m-2}a + \cdots + a^{m-1})$

2. Substitute the Factorized Form:

   After factorizing both the numerator and the denominator, we substitute them into the limit expression:

   $\lim_{x \to a} \frac{(x - a)(x^{n-1} + x^{n-2}a + \cdots + a^{n-1})}{(x - a)(x^{m-1} + x^{m-2}a + \cdots + a^{m-1})}$

3. Cancel Common Terms: The $(x - a)$ terms cancel out, leaving us with the following expression:

   $\lim_{x \to a} \frac{x^{n-1} + x^{n-2}a + \cdots + a^{n-1}}{x^{m-1} + x^{m-2}a + \cdots + a^{m-1}}$

4. Evaluate the Limit: Now, we can substitute $x = a$ directly into the remaining expression:

   $\frac{a^{n-1} + a^{n-1} + \cdots + a^{n-1}}{a^{m-1} + a^{m-1} + \cdots + a^{m-1}}$

   Both the numerator and denominator consist of sums of terms involving powers of $a$. The numerator has $n$ terms, each equal to $a^{n-1}$, so it simplifies to $n \cdot a^{n-1}$. Similarly, the denominator has $m$ terms, each equal to $a^{m-1}$, so it simplifies to $m \cdot a^{m-1}$.

5. Final Expression:

   The limit becomes:

   $\frac{n \cdot a^{n-1}}{m \cdot a^{m-1}}$

6. Simplify: This simplifies further to:

   $\frac{n}{m} \cdot a^{n-m}$

Final Answer:

$\lim_{x \to a} \frac{x^n - a^n}{x^m - a^m} = \frac{n}{m} \cdot a^{n-m}$

Let me know if you would like further clarifications.

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Here are five questions that build on this:

1. What happens if $n = m$ in the given limit?
2. How would you approach this limit if $a = 0$?
3. Can you explain the difference of powers formula and how it's used in limits?
4. How does the limit change if we were working with different variables, such as $x^p - a^p$ and $x^q - a^q$?
5. Could L'Hôpital's rule be used as an alternative method for solving this limit?

Tip: When factoring polynomials involving powers, always look for patterns like the difference of squares or sums of powers to simplify the expressions!