If `lim_( x rarr oo) [ (x^(3)+1)/(x^(2)+1)- (ax +b)]` =2 then

Options

A. a =1 and b =1
B. a=1 and b=-1
C. a=1 and b=-2
D. a=1 and b=2

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to \infty} \left( \frac{x^3 + 1}{x^2 + 1} - (ax + b) \right) = 2 \]

### Step 1: Rewrite the limit expression
We start by rewriting the expression inside the limit:

\[ \frac{x^3 + 1}{x^2 + 1} - (ax + b) \]

To combine these two terms, we will find a common denominator, which is \(x^2 + 1\):

\[ \frac{x^3 + 1 - (ax + b)(x^2 + 1)}{x^2 + 1} \]

### Step 2: Expand the numerator
Now we expand the numerator:

\[ x^3 + 1 - (ax^3 + bx^2 + ax + b) \]

This simplifies to:

\[ x^3 + 1 - ax^3 - bx^2 - ax - b \]

Combining like terms gives:

\[ (1 - a)x^3 - bx^2 - ax + (1 - b) \]

### Step 3: Set up the limit
Now we have:

\[ \lim_{x \to \infty} \frac{(1 - a)x^3 - bx^2 - ax + (1 - b)}{x^2 + 1} \]

### Step 4: Analyze the limit
As \(x\) approaches infinity, the highest degree term in the numerator will dominate. Thus, we focus on the term \((1 - a)x^3\).

The limit can be simplified to:

\[ \lim_{x \to \infty} \frac{(1 - a)x^3}{x^2} = \lim_{x \to \infty} (1 - a)x = \infty \quad \text{if } 1 - a \neq 0 \]

For the limit to equal 2, we need \(1 - a = 0\), which implies:

\[ a = 1 \]

### Step 5: Substitute \(a\) back into the limit
Now substituting \(a = 1\) back into the limit expression, we have:

\[ \lim_{x \to \infty} \frac{-bx^2 - x + (1 - b)}{x^2 + 1} \]

The leading term in the numerator is \(-bx^2\), so we analyze:

\[ \lim_{x \to \infty} \frac{-bx^2}{x^2} = -b \]

### Step 6: Set the limit equal to 2
For the limit to equal 2, we set:

\[ -b = 2 \implies b = -2 \]

### Conclusion
Thus, the values of \(a\) and \(b\) are:

\[ a = 1, \quad b = -2 \]

### Final Answer
The solution is \(a = 1\) and \(b = -2\).

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