It is given that f(x) = `(ax+b)/(x+1) , Lim _(x to 0) f(x) and Lim_(x to oo) f(x) = 1 ` Prove that fx(-2) = 0 .

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) in the function \( f(x) = \frac{ax + b}{x + 1} \) given the limits:

1. \( \lim_{x \to 0} f(x) = 1 \)
2. \( \lim_{x \to \infty} f(x) = 1 \)

Then, we will use these values to find \( f(-2) \) and prove that it equals 0.

### Step 1: Find \( \lim_{x \to \infty} f(x) \)

We start by calculating the limit as \( x \) approaches infinity:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax + b}{x + 1} \]

To simplify this, we divide the numerator and the denominator by \( x \):

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

As \( x \) approaches infinity, \( \frac{b}{x} \) and \( \frac{1}{x} \) both approach 0. Therefore, we have:

\[ = \frac{a + 0}{1 + 0} = a \]

Given that this limit equals 1, we have:

\[ a = 1 \]

### Step 2: Find \( \lim_{x \to 0} f(x) \)

Now we calculate the limit as \( x \) approaches 0:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{ax + b}{x + 1} \]

Substituting \( a = 1 \):

\[ = \lim_{x \to 0} \frac{1 \cdot x + b}{x + 1} = \lim_{x \to 0} \frac{x + b}{x + 1} \]

Substituting \( x = 0 \):

\[ = \frac{0 + b}{0 + 1} = b \]

Given that this limit also equals 1, we have:

\[ b = 1 \]

### Step 3: Find \( f(-2) \)

Now that we have \( a = 1 \) and \( b = 1 \), we can write the function:

\[ f(x) = \frac{1 \cdot x + 1}{x + 1} = \frac{x + 1}{x + 1} \]

Now we calculate \( f(-2) \):

\[ f(-2) = \frac{-2 + 1}{-2 + 1} = \frac{-1}{-1} = 1 \]

### Conclusion

We find that \( f(-2) = 1 \), not 0. Therefore, the statement to prove \( f(-2) = 0 \) is incorrect.