Question

If \(\underset{x\rightarrow \infty }{lim}[\frac{x^2+x+1}{x+1}-ax-b] = 3\), then \(a-b =\)...

• A. \(2\)
• B. \(3\)
• C. \(-2\)
• D. \(4\)

Hint:

Divide the polynomial first so the linear part can be matched.

Answer 1

The Correct Option is D

Step 1: Match coefficients:
Write $\frac{x^2 + x + 1}{x + 1} - ax - b = \frac{(1 - a)x^2 + (1 - a - b)x + (1 - b)}{x + 1}$.

Step 2: Conditions:
For the limit to be finite we need the $x^2$ term to vanish: $a = 1$. The limit is then the coefficient of $x$ over $x$: $-b = 3$, so $b = -3$.
So $a - b = 4$.

Final Answer:
$a - b = 4$, option (D). \[ \boxed{4} \]