The value of `lim_(ntooo)((a-1+root(n)(b))/a)^(n) (agt0,bgt0)` is equal to

Options

A. `root(a)(b)`
B. `root(b)(a)`
C. `sqrt(b)`
D. `sqrt(a)`

Answer

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

### Step 1: Rewrite the expression
We start by rewriting the limit in a more manageable form:
\[ L = \lim_{n \to \infty} \left( \frac{a - 1 + \sqrt[n]{b}}{a} \right)^n \]
This can be expressed as:
\[ L = \lim_{n \to \infty} \left( 1 + \frac{(a - 1) + \sqrt[n]{b} - a}{a} \right)^n \]
which simplifies to:
\[ L = \lim_{n \to \infty} \left( 1 + \frac{\sqrt[n]{b} - 1}{a} \right)^n \]

### Step 2: Analyze the term \( \sqrt[n]{b} \)
As \( n \to \infty \), we know that \( \sqrt[n]{b} \) approaches \( 1 \). Therefore, we can rewrite:
\[ L = \lim_{n \to \infty} \left( 1 + \frac{\sqrt[n]{b} - 1}{a} \right)^n \]

### Step 3: Use the exponential limit
This expression is of the form \( (1 + x_n)^n \) where \( x_n = \frac{\sqrt[n]{b} - 1}{a} \). We can use the fact that:
\[ \lim_{n \to \infty} (1 + x_n)^n = e^{\lim_{n \to \infty} n x_n} \]
if \( \lim_{n \to \infty} n x_n \) exists.

### Step 4: Find \( n x_n \)
Now we compute:
\[ n x_n = n \cdot \frac{\sqrt[n]{b} - 1}{a} \]
To evaluate \( \lim_{n \to \infty} n (\sqrt[n]{b} - 1) \), we can use the fact that \( \sqrt[n]{b} = e^{\frac{\log b}{n}} \). Thus:
\[ \sqrt[n]{b} - 1 = e^{\frac{\log b}{n}} - 1 \approx \frac{\log b}{n} \quad \text{(as \( n \to \infty \))} \]
This gives:
\[ n (\sqrt[n]{b} - 1) \approx n \cdot \frac{\log b}{n} = \log b \]

### Step 5: Substitute back into the limit
Now substituting back, we have:
\[ \lim_{n \to \infty} n x_n = \lim_{n \to \infty} \frac{\log b}{a} = \frac{\log b}{a} \]
Thus:
\[ L = e^{\frac{\log b}{a}} = b^{\frac{1}{a}} \]

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