Evaluate `lim_(xtooo)((x-3)/(x+2))^(x)`

Options

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

Answer

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

### Step 1: Rewrite the limit
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:
\[ \frac{x-3}{x+2} = \frac{x(1 - \frac{3}{x})}{x(1 + \frac{2}{x})} = \frac{1 - \frac{3}{x}}{1 + \frac{2}{x}} \]

### Step 2: Substitute into the limit
Now, substituting this back into the limit gives us:
\[ \lim_{x \to \infty} \left( \frac{1 - \frac{3}{x}}{1 + \frac{2}{x}} \right)^x \]

### Step 3: Identify the indeterminate form
As \( x \to \infty \), both \( \frac{3}{x} \) and \( \frac{2}{x} \) approach 0. Thus, the expression approaches:
\[ \frac{1 - 0}{1 + 0} = 1 \]
This leads us to the indeterminate form \( 1^\infty \).

### Step 4: Apply the limit property
For limits of the form \( \lim_{x \to a} f(x)^{g(x)} \) where \( f(x) \to 1 \) and \( g(x) \to \infty \), we can use the property:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) (f(x) - 1)} \]
In our case, let:
- \( f(x) = \frac{x-3}{x+2} \)
- \( g(x) = x \)

### Step 5: Calculate \( g(x)(f(x) - 1) \)
Now we need to compute:
\[ g(x) \left( f(x) - 1 \right) = x \left( \frac{x-3}{x+2} - 1 \right) \]
Calculating \( f(x) - 1 \):
\[ \frac{x-3}{x+2} - 1 = \frac{x-3 - (x+2)}{x+2} = \frac{-5}{x+2} \]
Thus,
\[ g(x) \left( f(x) - 1 \right) = x \cdot \frac{-5}{x+2} \]

### Step 6: Simplify the expression
Now, simplify:
\[ \frac{-5x}{x+2} \]
As \( x \to \infty \), this approaches:
\[ \frac{-5x}{x} = -5 \]

### Step 7: Substitute back into the limit
Now we substitute back into the limit:
\[ \lim_{x \to \infty} g(x) (f(x) - 1) = -5 \]
Thus, we have:
\[ \lim_{x \to \infty} \left( \frac{x-3}{x+2} \right)^x = e^{-5} \]

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