`lim_(x to a) (x^(n) -a^(n))/(x-a)`=

Options

A. `na^(n)`
B. `na^(n-1)`
C. 0
D. does not exist

Answer

To solve the limit \( \lim_{x \to a} \frac{x^n - a^n}{x - a} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Limit Form**:
We start with the limit:
\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} \]
When we substitute \( x = a \), we get:
\[ \frac{a^n - a^n}{a - a} = \frac{0}{0} \]
This is an indeterminate form.

**Hint**: Recognize that \( \frac{0}{0} \) is an indeterminate form, which indicates that we can apply L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
where \( f(x) = x^n - a^n \) and \( g(x) = x - a \).

3. **Differentiate the Numerator and Denominator**:
We differentiate the numerator and the denominator:
- The derivative of the numerator \( f(x) = x^n - a^n \) is:
\[ f'(x) = nx^{n-1} \]
- The derivative of the denominator \( g(x) = x - a \) is:
\[ g'(x) = 1 \]

4. **Re-evaluate the Limit**:
Now we can substitute back into the limit:
\[ \lim_{x \to a} \frac{nx^{n-1}}{1} \]
Substituting \( x = a \):
\[ = n a^{n-1} \]

5. **Final Result**:
Thus, we conclude that:
\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \]

### Summary:
The limit evaluates to:
\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \]