`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: Rewrite the expression
We start by rewriting the expression inside the limit:
\[ \frac{x+2}{x+1} = \frac{x(1 + \frac{2}{x})}{x(1 + \frac{1}{x})} = \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} \]
Thus, the limit can be rewritten as:
\[ \lim_{x \to \infty} \left(\frac{1 + \frac{2}{x}}{1 + \frac{1}{x}}\right)^{x+3} \]

### Step 2: Analyze the limit
As \( x \to \infty \), both \( \frac{2}{x} \) and \( \frac{1}{x} \) approach 0. Therefore:
\[ \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} \to \frac{1 + 0}{1 + 0} = 1 \]
This leads us to an indeterminate form \( 1^{\infty} \).

### Step 3: Apply the exponential limit
We can use the property of limits for indeterminate forms:
\[ \lim_{x \to \infty} f(x)^{g(x)} = e^{\lim_{x \to \infty} g(x) \cdot (f(x) - 1)} \]
Here, let:
- \( f(x) = \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} \)
- \( g(x) = x + 3 \)

### Step 4: Calculate \( f(x) - 1 \)
Now we need to find \( f(x) - 1 \):
\[ f(x) - 1 = \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} - 1 = \frac{(1 + \frac{2}{x}) - (1 + \frac{1}{x})}{1 + \frac{1}{x}} = \frac{\frac{2}{x} - \frac{1}{x}}{1 + \frac{1}{x}} = \frac{\frac{1}{x}}{1 + \frac{1}{x}} = \frac{1}{x(1 + \frac{1}{x})} \]

### Step 5: Substitute into the limit
Now we substitute \( f(x) - 1 \) into the limit:
\[ \lim_{x \to \infty} (x + 3) \cdot \left(\frac{1}{x(1 + \frac{1}{x})}\right) = \lim_{x \to \infty} \frac{x + 3}{x(1 + \frac{1}{x})} \]
This simplifies to:
\[ \lim_{x \to \infty} \frac{1 + \frac{3}{x}}{1 + \frac{1}{x}} = \frac{1 + 0}{1 + 0} = 1 \]

### Step 6: Final limit calculation
Now we have:
\[ \lim_{x \to \infty} g(x) \cdot (f(x) - 1) = 1 \]
Thus:
\[ \lim_{x \to \infty} \left(\frac{x+2}{x+1}\right)^{x+3} = e^1 = e \]

### Conclusion
The final result is:
\[ \lim_{x \to \infty} \left(\frac{x+2}{x+1}\right)^{x+3} = e \]