For ` x in R , lim_( x to infty) ((x -3)/(x + 2))^x ` is equal to

Options

A. e
B. `e^(-1)`
C. ` e^(-5)`
D. `e^(5)`

Explanation

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

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:
\[ \frac{x - 3}{x + 2} = \frac{x(1 - \frac{3}{x})}{x(1 + \frac{2}{x})} = \frac{1 - \frac{3}{x}}{1 + \frac{2}{x}} \]
Thus, we have:
\[ \lim_{x \to \infty} \left(\frac{x - 3}{x + 2}\right)^x = \lim_{x \to \infty} \left(\frac{1 - \frac{3}{x}}{1 + \frac{2}{x}}\right)^x \]

### Step 2: Apply the limit
Now we can apply the limit:
\[ \lim_{x \to \infty} \left(\frac{1 - \frac{3}{x}}{1 + \frac{2}{x}}\right)^x \]
As \( x \to \infty \), \( \frac{3}{x} \to 0 \) and \( \frac{2}{x} \to 0 \). Therefore, we can simplify:
\[ \lim_{x \to \infty} \left(\frac{1 - 0}{1 + 0}\right)^x = \lim_{x \to \infty} 1^x = 1 \]

### Step 3: Use the exponential limit
To evaluate the limit more rigorously, we can take the natural logarithm:
\[ \ln L = \lim_{x \to \infty} x \ln\left(\frac{1 - \frac{3}{x}}{1 + \frac{2}{x}}\right) \]
Using the logarithm properties, we can express this as:
\[ \ln L = \lim_{x \to \infty} x \left(\ln(1 - \frac{3}{x}) - \ln(1 + \frac{2}{x})\right) \]

### Step 4: Apply Taylor expansion
Using the Taylor expansion for \( \ln(1 + u) \) around \( u = 0 \):
\[ \ln(1 - u) \approx -u \quad \text{and} \quad \ln(1 + u) \approx u \]
we have:
\[ \ln(1 - \frac{3}{x}) \approx -\frac{3}{x} \quad \text{and} \quad \ln(1 + \frac{2}{x}) \approx \frac{2}{x} \]
Thus:
\[ \ln L = \lim_{x \to \infty} x \left(-\frac{3}{x} - \frac{2}{x}\right) = \lim_{x \to \infty} x \left(-\frac{5}{x}\right) = -5 \]

### Step 5: Exponentiate to find L
Now we can exponentiate to find \( L \):
\[ L = e^{-5} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} \left(\frac{x - 3}{x + 2}\right)^x = e^{-5} \]

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