If `lim _(xto oo) (sqrt( x ^(2) -x+1)-ax -b)=0,` then for `k ge 2, (k in N ) lim _(xto oo) sec ^(2n) (k ! pi b ) =`

Options

A. a
B. `-a`
C. 2a
D. b

Answer

To solve the limit problem step by step, we start with the given limit:

\[ \lim_{x \to \infty} \left( \sqrt{x^2 - x + 1} - ax - b \right) = 0 \]

### Step 1: Simplifying the Expression
We can rewrite the expression inside the limit:

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

Now, substituting this back into the limit gives:

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

### Step 2: Factor Out \(x\)
We can factor \(x\) out of the limit:

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

### Step 3: Finding the Limit
For the limit to equal 0 as \(x \to \infty\), the term inside the parentheses must approach 0. Thus, we need:

\[ \sqrt{1 - 0 + 0} - a = 0 \implies 1 - a = 0 \implies a = 1 \]

### Step 4: Finding \(b\)
Now substituting \(a = 1\) back into the limit gives:

\[ \lim_{x \to \infty} \left( x(1 - 1) - b \right) = -b \]

For this limit to equal 0, we must have \(b = 0\).

### Step 5: Conclusion for \(a\) and \(b\)
Thus, we have determined:

\[ a = 1 \quad \text{and} \quad b = 0 \]

### Step 6: Evaluating the Second Limit
Now we need to evaluate:

\[ \lim_{x \to \infty} \sec^{2n}(k! \pi b) \]

Substituting \(b = 0\):

\[ \lim_{x \to \infty} \sec^{2n}(k! \pi \cdot 0) = \sec^{2n}(0) = 1^{2n} = 1 \]

### Final Answer
Thus, the final answer is:

\[ \lim_{x \to \infty} \sec^{2n}(k! \pi b) = 1 \]