`lim_(xrarroo) ((x+2)/(x+1))^(x+3)` is equal to

Options

A. 1
B. e
C. `e^2`
D. `e^3`

Answer

To solve the limit \( \lim_{x \to \infty} \left( \frac{x+2}{x+1} \right)^{x+3} \), we can follow these steps:

### Step 1: Identify the Form
We first observe that as \( x \to \infty \), both the numerator and denominator approach infinity, leading to a form of \( \frac{\infty}{\infty} \). This means we can simplify the expression.

### Step 2: Simplify the Base
We simplify the base of the exponent:
\[ \frac{x+2}{x+1} = \frac{x(1 + \frac{2}{x})}{x(1 + \frac{1}{x})} = \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}}. \]
As \( x \to \infty \), \( \frac{2}{x} \to 0 \) and \( \frac{1}{x} \to 0 \), so:
\[ \frac{x+2}{x+1} \to \frac{1 + 0}{1 + 0} = 1. \]

### Step 3: Rewrite the Limit
Now we rewrite the limit in a form suitable for applying the exponential limit:
\[ \lim_{x \to \infty} \left( \frac{x+2}{x+1} \right)^{x+3} = \lim_{x \to \infty} e^{(x+3) \ln\left(\frac{x+2}{x+1}\right)}. \]

### Step 4: Find the Logarithm
Next, we need to find \( \ln\left(\frac{x+2}{x+1}\right) \):
\[ \ln\left(\frac{x+2}{x+1}\right) = \ln(x+2) - \ln(x+1). \]
Using the property of logarithms:
\[ \ln(x+2) - \ln(x+1) = \ln\left(\frac{x+2}{x+1}\right). \]

### Step 5: Apply L'Hôpital's Rule
We can analyze the limit of \( (x+3) \ln\left(\frac{x+2}{x+1}\right) \):
\[ \lim_{x \to \infty} (x+3) \left( \ln\left(1 + \frac{1}{x+1}\right) \right). \]
Using the approximation \( \ln(1 + u) \approx u \) for small \( u \):
\[ \ln\left(1 + \frac{1}{x+1}\right) \approx \frac{1}{x+1}. \]
Thus,
\[ \lim_{x \to \infty} (x+3) \cdot \frac{1}{x+1} = \lim_{x \to \infty} \frac{x+3}{x+1} = 1. \]

### Step 6: Final Limit Calculation
Now substituting back, we have:
\[ \lim_{x \to \infty} e^{(x+3) \ln\left(\frac{x+2}{x+1}\right)} = e^{1} = e. \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} \left( \frac{x+2}{x+1} \right)^{x+3} = e. \]