`Lim_(x->oo) ((x/(x+1))^a + sin (1/x))^x` is equal to

Options

A. `e^a-1`
B. `e^1-a`
C. `e`
D. `0`

Explanation

To solve the limit problem \( \lim_{x \to \infty} \left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right)^x \), we will break it down step by step.

### Step 1: Analyze the limit of each component as \( x \to \infty \)

1. **Evaluate \( \frac{x}{x+1} \)**:
\[ \frac{x}{x+1} = \frac{x}{x(1 + \frac{1}{x})} = \frac{1}{1 + \frac{1}{x}} \to 1 \quad \text{as } x \to \infty \]

2. **Evaluate \( \sin \left( \frac{1}{x} \right) \)**:
\[ \sin \left( \frac{1}{x} \right) \to \sin(0) = 0 \quad \text{as } x \to \infty \]

### Step 2: Substitute the limits into the expression

Now substituting these limits into the original expression:
\[ \left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right)^x \to \left( 1^a + 0 \right)^x = 1^x = 1 \]

### Step 3: Identify the indeterminate form

However, we need to be careful because we are dealing with the limit of the form \( (1 + 0)^x \) as \( x \to \infty \). This is an indeterminate form \( 1^\infty \).

### Step 4: Rewrite the expression

To resolve this, we can rewrite the expression using the exponential function:
\[ \lim_{x \to \infty} \left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right)^x = \lim_{x \to \infty} e^{x \ln\left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right)} \]

### Step 5: Find the limit of the logarithm

Now we need to evaluate:
\[ \lim_{x \to \infty} x \ln\left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right) \]

### Step 6: Simplify the logarithm

As \( x \to \infty \):
\[ \left( \frac{x}{x+1} \right)^a \to 1 \quad \text{and} \quad \sin \left( \frac{1}{x} \right) \to 0 \]
Thus,
\[ \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \to 1 + 0 = 1 \]

### Step 7: Expand the logarithm

Using the Taylor expansion for \( \ln(1 + u) \) where \( u \to 0 \):
\[ \ln\left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right) \approx \ln(1) + \frac{u}{1} = 0 + u \]

### Step 8: Evaluate the limit

We can now evaluate:
\[ x \ln\left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right) \to x \left( \left( \frac{x}{x+1} \right)^a - 1 + \sin \left( \frac{1}{x} \right) \right) \]

### Step 9: Final limit evaluation

As \( x \to \infty \):
\[ \left( \frac{x}{x+1} \right)^a \approx 1 - \frac{a}{x} \quad \text{and} \quad \sin \left( \frac{1}{x} \right) \approx \frac{1}{x} \]
Thus,
\[ x \left( -\frac{a}{x} + \frac{1}{x} \right) = -a + 1 \]

### Step 10: Conclude the limit

Finally, we have:
\[ \lim_{x \to \infty} e^{x \ln\left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right)} = e^{1 - a} \]

### Final Answer

The limit is:
\[ \lim_{x \to \infty} \left( \left( \frac{x}{x+1} \right)^a + \sin \left( \frac{1}{x} \right) \right)^x = e^{1-a} \]