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 problem, we need to find the limit:

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

### Step 1: Simplify the Expression
We start with the expression inside the limit:

\[ \sqrt{x^2 - x + 1} - ax - b \]

As \( x \to \infty \), we can approximate the square root:

\[ \sqrt{x^2 - x + 1} \approx \sqrt{x^2} = x \]

Thus, we rewrite the expression:

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

### Step 2: Set Up the Limit
Now we substitute this approximation back into our limit:

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

This simplifies to:

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

### Step 3: Determine Conditions for the Limit to Equal Zero
For the limit to equal zero, the coefficient of \( x \) must be zero:

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

Also, the constant term must equal zero:

\[ -\frac{1}{2} - b = 0 \implies b = -\frac{1}{2} \]

### Step 4: Substitute Values of \( a \) and \( b \)
Now we have \( a = 1 \) and \( b = -\frac{1}{2} \).

### Step 5: Evaluate the Second Limit
Next, we need to evaluate:

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

Substituting \( b = -\frac{1}{2} \):

\[ k! \pi b = k! \pi \left(-\frac{1}{2}\right) = -\frac{k! \pi}{2} \]

Now, we find:

\[ \sec^{2n}\left(-\frac{k! \pi}{2}\right) = \sec^{2n}\left(\frac{k! \pi}{2}\right) \]

### Step 6: Analyze the Value of \( \sec^{2n} \)
The secant function has periodic properties. Specifically, for \( k \geq 2 \):

- If \( k! \) is even, \( \sec\left(\frac{k! \pi}{2}\right) = \sec(0) = 1 \).
- If \( k! \) is odd, \( \sec\left(\frac{k! \pi}{2}\right) \) will be undefined.

Since \( k! \) is even for all \( k \geq 2 \):

\[ \sec^{2n}\left(-\frac{k! \pi}{2}\right) = 1^{2n} = 1 \]

### Final Answer
Thus, we conclude:

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