Consider the function `f(x)=((ax+1)/(bx+2))^(x)`, where `a,bgt0`, the `lim_(xtooo)f(x)` is

Options

A. exists for all values of a and b
B. zer for `altb`
C. non existent for `agtb`
D. `e^(-(1//a))` or `e^(-(1//b))` if `a=b`

Answer

To solve the limit of the function \( f(x) = \left( \frac{ax + 1}{bx + 2} \right)^x \) as \( x \) approaches infinity, we will follow these steps:

### Step 1: Analyze the function
The function is given by:
\[ f(x) = \left( \frac{ax + 1}{bx + 2} \right)^x \]
As \( x \) approaches infinity, both the numerator and denominator will be dominated by the terms involving \( x \).

### Step 2: Simplify the fraction
We can simplify the fraction inside the limit:
\[ \frac{ax + 1}{bx + 2} = \frac{a + \frac{1}{x}}{b + \frac{2}{x}} \]
As \( x \to \infty \), \( \frac{1}{x} \to 0 \) and \( \frac{2}{x} \to 0 \). Therefore, we have:
\[ \frac{ax + 1}{bx + 2} \to \frac{a}{b} \]

### Step 3: Substitute into the limit
Now substituting this back into the function gives us:
\[ f(x) \approx \left( \frac{a}{b} \right)^x \]

### Step 4: Evaluate the limit based on the value of \( \frac{a}{b} \)
1. **Case 1:** If \( a < b \), then \( \frac{a}{b} < 1 \). Therefore:
\[ \lim_{x \to \infty} f(x) = \left( \frac{a}{b} \right)^x \to 0 \]

2. **Case 2:** If \( a > b \), then \( \frac{a}{b} > 1 \). Therefore:
\[ \lim_{x \to \infty} f(x) = \left( \frac{a}{b} \right)^x \to \infty \]

3. **Case 3:** If \( a = b \), then \( \frac{a}{b} = 1 \). We have the indeterminate form \( 1^\infty \). We can use the exponential limit:
\[ f(x) = e^{x \ln\left(\frac{a}{b}\right)} \to e^{0} = 1 \]

### Final Result
Thus, we summarize the results:
- If \( a < b \), \( \lim_{x \to \infty} f(x) = 0 \)
- If \( a > b \), \( \lim_{x \to \infty} f(x) = \infty \)
- If \( a = b \), \( \lim_{x \to \infty} f(x) = 1 \)