Question

$\lim_{x \to \infty} \left[ \frac{x^2 + 1}{x + 1} - ax - b \right], \, (a, b \in \mathbb{R}) = 0$. Then

• A. a=0, b=1
• B. a= 1, b= -1
• C. a = -1, b = 1
• D. a = 0, b =0

Answer 1

The Correct Option is B

We are given the limit expression: \[ \lim_{x \to \infty} \left[ \frac{x^2 + 1}{x + 1} - ax - b \right] \]

Step 1: Simplify the rational expression
We start by simplifying the rational term \( \frac{x^2 + 1}{x + 1} \). Perform polynomial long division on \( \frac{x^2 + 1}{x + 1} \): Dividing \( x^2 + 1 \) by \( x + 1 \), we get: \[ \frac{x^2 + 1}{x + 1} = x - 1 + \frac{2}{x + 1} \] Thus, the original expression becomes: \[ \frac{x^2 + 1}{x + 1} - ax - b = \left( x - 1 + \frac{2}{x + 1} \right) - ax - b \] 

Step 2: Combine like terms
Now, combine like terms: \[ x - 1 + \frac{2}{x + 1} - ax - b = (x - ax) + (-1 - b) + \frac{2}{x + 1} \] \[ = (1 - a)x + (-1 - b) + \frac{2}{x + 1} \] 

Step 3: Evaluate the limit
As \( x \to \infty \), the term \( \frac{2}{x + 1} \to 0 \), so the expression simplifies to: \[ (1 - a)x + (-1 - b) \] For the limit to be 0, the coefficient of \( x \) must be 0, i.e., \( 1 - a = 0 \), which gives \( a = 1 \). Additionally, the constant term must also be 0, i.e., \( -1 - b = 0 \), which gives \( b = -1 \). 

Conclusion:
Therefore, the values of \( a \) and \( b \) are \( a = 1 \) and \( b = -1 \).

Answer:

\[ \boxed{a = 1, b = -1} \]