The values of constants a and b so that
`lim_(x to infty) ((x^(2)+1)/(x+1)- ax -b)=0`, are

Options

A. a=0, b=0
B. `a=1, b=-1`
C. `a=-1, b=1`
D. `a=2, b=-1`

Answer

To solve the limit problem, we need to find the values of constants \( a \) and \( b \) such that:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - ax - b \right) = 0 \]

### Step 1: Simplify the expression inside the limit

We start with the expression:

\[ \frac{x^2 + 1}{x + 1} - ax - b \]

To combine the terms, we first find a common denominator:

\[ \frac{x^2 + 1 - (ax + b)(x + 1)}{x + 1} \]

### Step 2: Expand the numerator

Now, we expand the numerator:

\[ x^2 + 1 - (ax^2 + ax + bx + b) = x^2 + 1 - ax^2 - (a + b)x - b \]

This simplifies to:

\[ (1 - a)x^2 + (1 - (a + b))x + (1 - b) \]

### Step 3: Rewrite the limit

Now we can rewrite the limit as:

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

### Step 4: Analyze the limit

For the limit to equal 0 as \( x \to \infty \), the coefficient of \( x^2 \) in the numerator must be zero:

\[ 1 - a = 0 \implies a = 1 \]

### Step 5: Substitute \( a \) back into the limit

Now substituting \( a = 1 \) into the expression, we have:

\[ \lim_{x \to \infty} \frac{(1 - (1 + b))x + (1 - b)}{x + 1} \]

This simplifies to:

\[ \lim_{x \to \infty} \frac{-bx + (1 - b)}{x + 1} \]

### Step 6: Set the coefficient of \( x \) to zero

For the limit to be finite (specifically 0), the coefficient of \( x \) must also be zero:

\[ -b = 0 \implies b = -1 \]

### Conclusion

Thus, the values of the constants are:

\[ a = 1, \quad b = -1 \]

### Final Answer

The values of constants \( a \) and \( b \) are \( a = 1 \) and \( b = -1 \).

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