`lim_(xtooo)((x+1)/(x+2))^(2x+1)` is

Options

A. e
B. `e^(-2)`
C. `e^(-1)`
D. 1

Answer

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

### Step 1: Rewrite the Limit
We start by rewriting the limit in a more manageable form:
\[ L = \lim_{x \to \infty} \left( \frac{x+1}{x+2} \right)^{2x+1} \]

### Step 2: Identify the Indeterminate Form
As \( x \to \infty \), both the numerator and denominator approach infinity, leading to the form \( \frac{\infty}{\infty} \). Thus, we can express this limit using the exponential function:
\[ L = e^{\lim_{x \to \infty} (2x+1) \ln\left(\frac{x+1}{x+2}\right)} \]

### Step 3: Simplify the Logarithm
Next, we simplify the logarithm:
\[ \ln\left(\frac{x+1}{x+2}\right) = \ln(x+1) - \ln(x+2) \]
Using properties of logarithms, we can express this as:
\[ \ln\left(\frac{x+1}{x+2}\right) = \ln\left(1 - \frac{1}{x+2}\right) \]

### Step 4: Use Taylor Expansion
For large \( x \), we can use the Taylor expansion \( \ln(1 - u) \approx -u \) when \( u \) is small:
\[ \ln\left(1 - \frac{1}{x+2}\right) \approx -\frac{1}{x+2} \]

### Step 5: Substitute Back into the Limit
Now substituting back into our limit:
\[ L = e^{\lim_{x \to \infty} (2x+1) \left(-\frac{1}{x+2}\right)} \]

### Step 6: Simplify the Expression
We can simplify the expression:
\[ (2x + 1) \left(-\frac{1}{x + 2}\right) = -\frac{2x + 1}{x + 2} \]
Now, we can divide the numerator and denominator by \( x \):
\[ -\frac{2 + \frac{1}{x}}{1 + \frac{2}{x}} \]

### Step 7: Evaluate the Limit
Taking the limit as \( x \to \infty \):
\[ \lim_{x \to \infty} -\frac{2 + \frac{1}{x}}{1 + \frac{2}{x}} = -\frac{2 + 0}{1 + 0} = -2 \]

### Step 8: Final Result
Now substituting back into our expression for \( L \):
\[ L = e^{-2} \]

Thus, the final answer is:
\[ \lim_{x \to \infty} \left( \frac{x+1}{x+2} \right)^{2x+1} = e^{-2} \]

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