Let `y(x)=ax^(n)anddeltay` dentoe samll change in y. what is limit of `(deltay)/(deltax)asdeltaxrarr0`?

Options

A. 0
B. 1
C. `anx^(n-1)`
D. `ax^(n)log(ax)`

Explanation

To solve the problem, we need to find the limit of the ratio of the small change in \( y \) (denoted as \( \Delta y \)) to the small change in \( x \) (denoted as \( \Delta x \)) as \( \Delta x \) approaches 0. The function given is \( y(x) = ax^n \).

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function:
\[ y(x) = ax^n \]

2. **Calculate the Change in \( y \)**:
The small change in \( y \) can be expressed as:
\[ \Delta y = y(x + \Delta x) - y(x) = a(x + \Delta x)^n - ax^n \]

3. **Expand \( (x + \Delta x)^n \)**:
Using the binomial expansion, we can expand \( (x + \Delta x)^n \):
\[ (x + \Delta x)^n = x^n + n x^{n-1} \Delta x + \frac{n(n-1)}{2} x^{n-2} (\Delta x)^2 + \ldots \]
Therefore, we have:
\[ \Delta y = a \left( x^n + n x^{n-1} \Delta x + \frac{n(n-1)}{2} x^{n-2} (\Delta x)^2 + \ldots \right) - ax^n \]
Simplifying this gives:
\[ \Delta y = a \left( n x^{n-1} \Delta x + \frac{n(n-1)}{2} x^{n-2} (\Delta x)^2 + \ldots \right) \]

4. **Divide by \( \Delta x \)**:
Now, we divide \( \Delta y \) by \( \Delta x \):
\[ \frac{\Delta y}{\Delta x} = a \left( n x^{n-1} + \frac{n(n-1)}{2} x^{n-2} \Delta x + \ldots \right) \]

5. **Take the Limit as \( \Delta x \to 0 \)**:
Now, we take the limit as \( \Delta x \) approaches 0:
\[ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = a \left( n x^{n-1} + 0 + \ldots \right) = a n x^{n-1} \]

6. **Evaluate at \( x = 0 \)**:
Finally, we evaluate this limit at \( x = 0 \):
\[ \lim_{x \to 0} a n x^{n-1} = a n \cdot 0^{n-1} \]
If \( n > 1 \), this limit is 0. If \( n = 1 \), it is \( a \). If \( n < 1 \), it diverges.

### Conclusion:
Thus, the limit of \( \frac{\Delta y}{\Delta x} \) as \( \Delta x \to 0 \) is:
\[ \begin{cases} 0 & \text{if } n > 1 \\ a & \text{if } n = 1 \\ \text{undefined} & \text{if } n < 1 \end{cases} \]