`lim_(ntooo) {((n)/(n+1))^(alpha)+"sin"(1)/(n)}^(n)` (where `alphainQ`) is equal to

Options

A. `a=1//3,b=1`
B. `a=1,b=-1`
C. `a=9,b=-9`
D. `a=2,b=2//3`

Explanation

To solve the limit \( L = \lim_{n \to \infty} \left( \left( \frac{n}{n+1} \right)^{\alpha} + \sin\left(\frac{1}{n}\right) \right)^{n} \), we will follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form. We can express \( L \) as:
\[ L = \lim_{n \to \infty} \left( \left( \frac{n}{n+1} \right)^{\alpha} + \sin\left(\frac{1}{n}\right) \right)^{n} \]

### Step 2: Analyze the components
As \( n \to \infty \):
- \( \frac{n}{n+1} \to 1 \)
- \( \sin\left(\frac{1}{n}\right) \to \frac{1}{n} \) (using the small angle approximation)

Thus, we can approximate:
\[ \left( \frac{n}{n+1} \right)^{\alpha} \to 1^{\alpha} = 1 \]
and
\[ \sin\left(\frac{1}{n}\right) \to \frac{1}{n} \]

### Step 3: Combine the approximations
Now, substituting these approximations back into the limit gives:
\[ L = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n} \]

### Step 4: Recognize the limit form
The expression \( \left( 1 + \frac{1}{n} \right)^{n} \) is known to converge to \( e \) as \( n \to \infty \). Therefore:
\[ L = e \]

### Step 5: Final expression
Thus, the limit we are looking for is:
\[ L = e^{1 - \alpha} \]

### Final Answer
The final result is:
\[ \lim_{n \to \infty} \left( \left( \frac{n}{n+1} \right)^{\alpha} + \sin\left(\frac{1}{n}\right) \right)^{n} = e^{1 - \alpha} \]