Question

If \(\lim _{x\rightarrow \infty }\left[\frac{x^{3}+1}{x^{2}+1}-(ax+b)\right]=2\) then

• A. \(a=1,\ b=1\)
• B. \(a=1,\ b=-1\)
• C. \(a=1,\ b=-2\)
• D. \(a=1,\ b=2\)

Hint:

Combine into one fraction and make the highest power in the numerator vanish to get a, then match the limit to find b.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For the limit to be finite, the numerator must have a degree no higher than the denominator after combining terms.

Step 2: Combine
\(\dfrac{x^3+1}{x^2+1}-(ax+b)=\dfrac{(1-a)x^3-bx^2-ax+(1-b)}{x^2+1}\).

Step 3: Find a
The \(x^3\) term must vanish, so \(1-a=0\) and \(a=1\).

Step 4: Find b
The limit becomes \(\lim\dfrac{-bx^2-x+(1-b)}{x^2+1}=-b\). Setting \(-b=2\) gives \(b=-2\).

Final Answer:
The values are a = 1 and b = -2. \[ \boxed{\text{(C) }a=1,\ b=-2} \]