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: Understanding the Concept:
For the limit to be finite as \(x \to \infty\), the coefficient of \(x\) in the expression must vanish. Simplify the fraction by division.

Step 2: Key Formula or Approach:
\(\frac{x^2 + x + 1}{x + 1} = x + \frac{1}{x + 1}\), since \(x(x+1) = x^2 + x\) leaves remainder \(1\).

Step 3: Detailed Explanation:
The expression becomes \((1 - a)x - b + \frac{1}{x+1}\).
As \(x \to \infty\), \(\frac{1}{x+1} \to 0\). For a finite limit, \(1 - a = 0\), so \(a = 1\).
Then the limit is \(-b = 3\), so \(b = -3\).
\[ a - b = 1 - (-3) = 4 \]

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