If `lim_(xtooo) ((x^(2)+x+1)/(x+1)-ax-b)=4,` then

Explanation

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

\[ \lim_{x \to \infty} \left( \frac{x^2 + x + 1}{x + 1} - ax - b \right) = 4 \]

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

Start with the expression:

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

We can combine the terms by finding a common denominator. The common denominator is \( x + 1 \):

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

### Step 2: Expand the numerator

Now, we need to expand the numerator:

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

This simplifies to:

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

### Step 3: Combine like terms

Combine the like terms in the numerator:

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

So we have:

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

### Step 4: Divide by \( x \)

Now, divide the numerator and the denominator by \( x \):

\[ \frac{(1 - a)x + (1 - a - b) + \frac{1 + b}{x}}{1 + \frac{1}{x}} \]

### Step 5: Take the limit as \( x \to \infty \)

As \( x \to \infty \), the terms \( \frac{1 + b}{x} \) and \( \frac{1}{x} \) will approach 0. Thus, we are left with:

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

For the limit to exist and equal to 4, the coefficient of \( x \) must be 0:

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

### Step 6: Substitute \( a \) back to find \( b \)

Now substitute \( a = 1 \) into the remaining expression:

\[ 1 - 1 - b = 4 \implies -b = 4 \implies b = -4 \]

### Conclusion

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

\[ a = 1, \quad b = -4 \]