`("limit")_(x to a) (x^(m) -a^(m))/(x^(n)-a^(n))` is:

Options

A. `m/n a^(m-n)`
B. `n/m a^(n-m)`
C. `mna^(mn)`
D. none

Answer

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

### Step 1: Identify the form of the limit
When we substitute \(x = a\) directly into the expression, we get:
\[ \frac{a^m - a^m}{a^n - a^n} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

The numerator is \(x^m - a^m\) and the denominator is \(x^n - a^n\).

Taking the derivatives:
- Derivative of the numerator:
\[ \frac{d}{dx}(x^m - a^m) = mx^{m-1} \]
- Derivative of the denominator:
\[ \frac{d}{dx}(x^n - a^n) = nx^{n-1} \]

### Step 3: Rewrite the limit using derivatives
Now we can rewrite the limit as:
\[ \lim_{x \to a} \frac{mx^{m-1}}{nx^{n-1}} \]

### Step 4: Substitute \(x = a\) into the new limit
Now, substituting \(x = a\) into the limit gives:
\[ \frac{ma^{m-1}}{na^{n-1}} = \frac{m}{n} \cdot \frac{a^{m-1}}{a^{n-1}} = \frac{m}{n} \cdot a^{(m-n)} \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to a} \frac{x^m - a^m}{x^n - a^n} = \frac{m}{n} a^{m-n} \]

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