If `agt0, b gt0` than `lim_(nrarroo) ((a-1+b^((1)/(n)))/(a))^(n)=`

Options

A. `b^((1)/(a))`
B. `a^((1)/(b))`
C. `a^(b)`
D. `b^(a)`

Explanation

To solve the limit problem \( \lim_{n \to \infty} \left( \frac{a - 1 + b^{\frac{1}{n}}}{a} \right)^{n} \) where \( a > 0 \) and \( b > 0 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ L = \lim_{n \to \infty} \left( \frac{a - 1 + b^{\frac{1}{n}}}{a} \right)^{n} \]
This can be rewritten as:
\[ L = \lim_{n \to \infty} \left( 1 + \frac{b^{\frac{1}{n}} - 1}{a} \right)^{n} \]

### Step 2: Identify the indeterminate form
As \( n \to \infty \), \( b^{\frac{1}{n}} \) approaches 1, leading to the form \( 1^{\infty} \). To resolve this, we can use the exponential limit form.

### Step 3: Use the exponential limit
We can use the fact that:
\[ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^{n} = e^{x} \]
Thus, we need to express our limit in this form. We rewrite:
\[ L = \lim_{n \to \infty} e^{n \cdot \frac{b^{\frac{1}{n}} - 1}{a}} \]

### Step 4: Simplify the exponent
Next, we need to evaluate:
\[ \lim_{n \to \infty} n \cdot \frac{b^{\frac{1}{n}} - 1}{a} \]
Using the property that \( b^{\frac{1}{n}} - 1 \) can be approximated by \( \frac{\log b}{n} \) as \( n \to \infty \), we have:
\[ b^{\frac{1}{n}} - 1 \approx \frac{\log b}{n} \]
Thus:
\[ n \cdot (b^{\frac{1}{n}} - 1) \approx n \cdot \frac{\log b}{n} = \log b \]

### Step 5: Substitute back into the limit
Now substituting back, we get:
\[ L = e^{\frac{\log b}{a}} = b^{\frac{1}{a}} \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{n \to \infty} \left( \frac{a - 1 + b^{\frac{1}{n}}}{a} \right)^{n} = b^{\frac{1}{a}} \]