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 analyze three cases based on the relationship between \( a \) and \( b \).

### Step 1: Identify the cases based on \( a \) and \( b \)

We have three cases to consider:
1. Case 1: \( a < b \)
2. Case 2: \( a > b \)
3. Case 3: \( a = b \)

### Step 2: Case 1 - \( a < b \)

In this case, we analyze the limit:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( \frac{ax + 1}{bx + 2} \right)^x \]

Dividing the numerator and denominator by \( x \):

\[ = \lim_{x \to \infty} \left( \frac{a + \frac{1}{x}}{b + \frac{2}{x}} \right)^x \]

As \( x \to \infty \), \( \frac{1}{x} \) and \( \frac{2}{x} \) approach 0:

\[ = \lim_{x \to \infty} \left( \frac{a}{b} \right)^x \]

Since \( a < b \), \( \frac{a}{b} < 1 \). Therefore, \( \left( \frac{a}{b} \right)^x \to 0 \) as \( x \to \infty \).

### Step 3: Case 2 - \( a > b \)

Now, we analyze the limit for the case where \( a > b \):

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( \frac{ax + 1}{bx + 2} \right)^x \]

Again, dividing by \( x \):

\[ = \lim_{x \to \infty} \left( \frac{a + \frac{1}{x}}{b + \frac{2}{x}} \right)^x \]

As \( x \to \infty \):

\[ = \lim_{x \to \infty} \left( \frac{a}{b} \right)^x \]

Since \( a > b \), \( \frac{a}{b} > 1 \). Therefore, \( \left( \frac{a}{b} \right)^x \to \infty \) as \( x \to \infty \).

### Step 4: Case 3 - \( a = b \)

In this case, we have:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( \frac{ax + 1}{bx + 2} \right)^x \]

Since \( a = b \):

\[ = \lim_{x \to \infty} \left( \frac{a + \frac{1}{x}}{a + \frac{2}{x}} \right)^x \]

This simplifies to:

\[ = \lim_{x \to \infty} \left( \frac{a}{a} \right)^x = 1^x \]

This form \( 1^\infty \) is indeterminate. To resolve this, we can use the logarithmic limit:

\[ \lim_{x \to \infty} x \left( \frac{a + \frac{1}{x}}{a + \frac{2}{x}} - 1 \right) \]

Calculating the expression inside the limit:

\[ = \lim_{x \to \infty} x \left( \frac{a + \frac{1}{x} - (a + \frac{2}{x})}{a + \frac{2}{x}} \right) = \lim_{x \to \infty} x \left( \frac{-\frac{1}{x}}{a + \frac{2}{x}} \right) \]

This simplifies to:

\[ = \lim_{x \to \infty} \frac{-1}{a + \frac{2}{x}} = \frac{-1}{a} \]

Thus, we have:

\[ \lim_{x \to \infty} f(x) = e^{-\frac{1}{a}} \]

### Conclusion

The limit of \( f(x) \) as \( x \to \infty \) is:

- \( 0 \) if \( a < b \)
- \( \infty \) if \( a > b \)
- \( e^{-\frac{1}{a}} \) if \( a = b \)