We are given: $\underset{x\to \infty }{\lim }(\frac{x^2+x+1}{x+1}-ax-b)=4$. We need to find values of a and b.

Step 1: Simplify the expression inside the limit.

Let L = $\underset{x\to \infty }{\lim }(\frac{x^2+x+1}{x+1}-ax-b)$

First, perform polynomial division on the fraction:

$\frac{x^2+x+1}{x+1}=x+\frac{1}{x+1}$

So the expression becomes:

L = $\underset{x\to \infty }{\lim }(x+\frac{1}{x+1}-ax-b)$

Combine the x terms:

L = $\underset{x\to \infty }{\lim }\left(\right(1-a)x+\frac{1}{x+1}-b)$

Step 2: Analyze the limit as x approaches infinity.

For the limit to exist and be finite (equal to 4), the coefficient of the x term (which grows without bound) must be zero. Otherwise, the limit would be ±∞.

Therefore, we must have: $1-a=0$, which gives $a=1$.

Step 3: Substitute a=1 and evaluate the remaining limit.

Now the expression becomes:

L = $\underset{x\to \infty }{\lim }\left(\right(0)x+\frac{1}{x+1}-b)=\underset{x\to \infty }{\lim }(\frac{1}{x+1}-b)$

As $x\to \infty$, $\frac{1}{x+1}\to 0$.

Therefore, L = $0-b=-b$

We are given that this limit equals 4:

$-b=4$

Thus, $b=-4$

Final Answer: a = 1, b = -4

This corresponds to the option: a = 1, b = – 4

Related Concepts and Theory

Limits at Infinity: When evaluating the limit of a rational function (a ratio of polynomials) as x approaches infinity, the behavior is determined by the degrees of the numerator and denominator polynomials.

• If the degree of the numerator is greater than the degree of the denominator, the limit is ±∞.
• If the degree of the numerator is equal to the degree of the denominator, the limit is the ratio of the leading coefficients.
• If the degree of the numerator is less than the degree of the denominator, the limit is 0.

In this problem, we had an expression that simplified to a linear term in x plus other terms. For the overall limit to be a finite number, the coefficient of this unbounded term must be zero.

Key Formulae

Polynomial Division: For dividing a polynomial P(x) by (x - c), or in this case, (x + 1).

Limit of 1/x: $\underset{x\to \infty }{\lim }\frac{1}{x}=0$

General Strategy for limits of the form: $\underset{x\to \infty }{\lim }\left[f\right(x)-(ax+b\left)\right]=L$ is to ensure the terms that grow with x cancel out, leaving only finite terms.