`lim_(x -> oo) ((x-3)/(x+2))^x` is equal to

Options

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

Answer

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

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to \infty} \left( \frac{x-3}{x+2} \right)^x \]
We can simplify the fraction 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 can rewrite the limit as:
\[ \lim_{x \to \infty} \left( \frac{1 - \frac{3}{x}}{1 + \frac{2}{x}} \right)^x \]

### Step 2: Evaluate the limit of the base
As \( x \to \infty \), both \( \frac{3}{x} \) and \( \frac{2}{x} \) approach 0. Therefore:
\[ \frac{1 - \frac{3}{x}}{1 + \frac{2}{x}} \to \frac{1 - 0}{1 + 0} = 1 \]

### Step 3: Use the exponential limit
Since the base approaches 1, we can use the fact that \( (1 + u)^n \approx e^{nu} \) for small \( u \). We need to express the limit in a suitable form:
\[ \lim_{x \to \infty} \left( \frac{1 - \frac{3}{x}}{1 + \frac{2}{x}} \right)^x = \lim_{x \to \infty} \left( 1 - \frac{3}{x} - \frac{2}{x} + O\left(\frac{1}{x^2}\right) \right)^x \]
This simplifies to:
\[ \lim_{x \to \infty} \left( 1 - \frac{5}{x} + O\left(\frac{1}{x^2}\right) \right)^x \]

### Step 4: Apply the limit
Using the exponential limit:
\[ \lim_{x \to \infty} \left( 1 - \frac{5}{x} \right)^x = e^{-5} \]

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

### Summary
The final answer is:
\[ \boxed{e^{-5}} \]