If `lim_(xtooo)((x^(2)+1)/(x+1)-ax-b)=0`, find the values of a and b.

Answer

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

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

### Step-by-Step Solution:

**Step 1: Combine the terms into a single fraction.**

We start with the expression inside the limit:

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

To combine these terms, we can rewrite it as:

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

**Step 2: Expand the numerator.**

Now, we need to expand the numerator:

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

This simplifies to:

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

**Step 3: Set up the limit.**

Now we have:

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

**Step 4: Analyze the degrees of the numerator and denominator.**

For the limit to exist and equal zero as \( x \to \infty \), the degree of the numerator must be less than the degree of the denominator. The denominator has a degree of 1, so we need the coefficient of \( x^2 \) in the numerator to be zero:

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

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

Now substituting \( a = 1 \) into the numerator:

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

So the limit becomes:

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

**Step 6: Simplify the limit expression.**

Now we can simplify this limit:

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

**Step 7: Set the limit equal to zero.**

For the limit to equal zero:

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

### Final Values:

Thus, the values of \( a \) and \( b \) are:

\[ \boxed{a = 1, b = 1} \]