The value of the constant `alpha` and `beta` such that `lim_(x rarr oo) ((x^(2)+1)/(x+1) - alpha x - beta)` =0 are respectively

Options

A. (1,1)
B. `(-1,1)`
C. `(1,-1)`
D. (0,1)

Answer

To solve the problem, we need to find the constants \(\alpha\) and \(\beta\) such that:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0 \]

### Step 1: Simplify the expression inside the limit

First, we rewrite the expression inside the limit:

\[ \frac{x^2 + 1}{x + 1} - \alpha x - \beta \]

To combine the fractions, we need a common denominator, which is \(x + 1\):

\[ = \frac{x^2 + 1 - \alpha x (x + 1) - \beta (x + 1)}{x + 1} \]

### Step 2: Expand the numerator

Now, we expand the numerator:

\[ = x^2 + 1 - \alpha x^2 - \alpha x - \beta x - \beta \]

Combining like terms gives us:

\[ = (1 - \alpha)x^2 + (1 - \alpha - \beta)x + (1 - \beta) \]

### Step 3: Set the limit condition

For the limit to approach 0 as \(x\) approaches infinity, the degree of the numerator must be less than the degree of the denominator. The denominator is of degree 1, so we need the coefficient of \(x^2\) in the numerator to be 0:

\[ 1 - \alpha = 0 \implies \alpha = 1 \]

### Step 4: Solve for \(\beta\)

Now, substituting \(\alpha = 1\) into the linear term of the numerator:

\[ 1 - \alpha - \beta = 0 \implies 1 - 1 - \beta = 0 \implies -\beta = 0 \implies \beta = 0 \]

### Step 5: Check the constant term

Finally, we check the constant term:

\[ 1 - \beta = 0 \implies 1 - 0 = 1 \quad \text{(This does not affect the limit)} \]

### Conclusion

Thus, the values of \(\alpha\) and \(\beta\) are:

\[ \alpha = 1, \quad \beta = 0 \]