`lim_(x to infty) ((x+2)/(x+1))^(x+3)` =

Options

A. e
B. `e^(2)`
C. `e^(4)`
D. `e^(6)`

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 with the limit:
\[ L = \lim_{x \to \infty} \left( \frac{x+2}{x+1} \right)^{x+3} \]

### Step 2: Simplify the base
We can simplify the base \( \frac{x+2}{x+1} \):
\[ \frac{x+2}{x+1} = \frac{x(1 + \frac{2}{x})}{x(1 + \frac{1}{x})} = \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} \]
As \( x \to \infty \), both \( \frac{2}{x} \) and \( \frac{1}{x} \) approach 0. Thus:
\[ \frac{x+2}{x+1} \to \frac{1 + 0}{1 + 0} = 1 \]

### Step 3: Identify the indeterminate form
Now, we have:
\[ L = \lim_{x \to \infty} \left( 1 \right)^{x+3} \]
This is an indeterminate form of type \( 1^\infty \).

### Step 4: Use logarithmic transformation
To resolve the indeterminate form, we can take the natural logarithm:
\[ \ln L = \lim_{x \to \infty} (x+3) \ln \left( \frac{x+2}{x+1} \right) \]

### Step 5: Simplify the logarithm
Next, we simplify \( \ln \left( \frac{x+2}{x+1} \right) \):
\[ \ln \left( \frac{x+2}{x+1} \right) = \ln(x+2) - \ln(x+1) \]
Using properties of logarithms:
\[ \ln(x+2) - \ln(x+1) = \ln \left( \frac{x+2}{x+1} \right) = \ln \left( 1 + \frac{1}{x+1} \right) \]
For large \( x \), we can use the approximation \( \ln(1 + u) \approx u \) when \( u \) is small:
\[ \ln \left( 1 + \frac{1}{x+1} \right) \approx \frac{1}{x+1} \]

### Step 6: Substitute back into the limit
Now substituting back:
\[ \ln L = \lim_{x \to \infty} (x+3) \cdot \frac{1}{x+1} \]
This simplifies to:
\[ \ln L = \lim_{x \to \infty} \frac{x+3}{x+1} \]
As \( x \to \infty \), this approaches:
\[ \ln L = \lim_{x \to \infty} \frac{1 + \frac{3}{x}}{1 + \frac{1}{x}} = \frac{1 + 0}{1 + 0} = 1 \]

### Step 7: Exponentiate to find L
Now, we exponentiate both sides:
\[ L = e^{\ln L} = e^1 = e \]

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