Question

If \(\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x-b\right)=0\), then for \(k \geq 2, k \in\) \(\mathrm{N}\) which of the following is/are correct ? (A) \(2 \mathrm{a}+\mathrm{b}=0\) (B) \(a+2 b=0\) (C) \(\lim _{n \rightarrow \infty} \sec ^{2 n}(k ! \pi b)=1\) (D) \(\lim _{n \rightarrow \infty} \sec ^{2 n}(k ! \pi a)=1\)

Short answer

Based on the step-by-step solution, determine the values of a and b so that \(\lim_{x \to \infty}(\sqrt{x^2 - x + 1} - ax - b) = 0\), and verify which options are correct. After performing an analysis and simplifying the given expression, we find out that the values for a and b are: \(a = 1\) and \(b = 1\). Given this, we can verify which options related to the value of k are correct. The correct options are: (C) \(\lim_{n\to\infty} \sec^{2n}(k!\pi b) = \lim_{n\to\infty} \sec^{2n}(k! \pi (1)) = \lim_{n\to\infty} \sec^{2n}(k!\pi)\) which equals 1. (D) \(\lim_{n\to\infty} \sec^{2n}(k!\pi a) = \lim_{n\to\infty} \sec^{2n}(k! \pi (1)) = \lim_{n\to\infty} \sec^{2n}(k!\pi)\) which also equals 1.

Step-by-step solution

Simplify the expression within the limit

Let's start by simplifying the expression within the limit, \(\sqrt{x^2 - x + 1} - ax - b\). To do this, we will manipulate the expression to obtain common terms and easier expressions to work with. Multiply and divide by the conjugate of the given expression, which is \(\sqrt{x^2 - x + 1} + ax + b\): $$\lim_{x \to \infty} \frac{x^2 - x + 1 - (ax + b)^2}{\sqrt{x^2 - x + 1} + ax + b}$$

Simplify the numerator

Now, let's simplify the numerator of the expression. Expand the square of \((ax+b)\): \(x^2 - x + 1 - (a^2 x^2 + 2abx + b^2)\). Further simplifying, we get: $$\lim_{x \to \infty} \frac{(1-a^2)x^2 - (2ab+1)x + (1-b^2)}{\sqrt{x^2 - x + 1} + ax + b}$$

Apply the limit

As \(x \to \infty\), only the terms with the highest power of x will contribute to the limit. Therefore, we can rewrite the limit as: $$\lim_{x \to \infty} \frac{(1 - a^2)x^2}{x(\sqrt{x^2 - x + 1} + ax + b)}$$ Divide both the numerator and denominator by \(x\): $$\lim_{x \to \infty} \frac{(1 - a^2)x}{\sqrt{x^2 - x + 1} + ax + b} = 0$$ Since the limit is equal to 0, we need to find the values of a and b that satisfy this condition.

Find a and b

Observe that the limit equals 0 when the degree of the numerator and denominator is equal. From the limit expression, we can conclude that: \(a^2 = 1\). But since \(a\neq0\) (if \(a=0\), denominator will be too small and the fraction will go undefined), then \(a=1\). Now we need to find \(b\). Set the denominator equal to \(x\) and solve for \(b\): $$\sqrt{x^2 - x + 1} + x = x + b$$ By subtracting \(x\) from both sides, we obtain: $$\sqrt{x^2 - x + 1} - x = -b$$ So, by comparing our expression with the problem definition, it implies \(b = 1\).

Verify each option

Now that we have the values for a and b, we can check each option: (A) \(2a + b = 0 \implies 2(1) + 1 = 3 \neq 0\). So, this option is incorrect. (B) \(a + 2b = 0 \implies 1 + 2(1) = 3 \neq 0\). So, this option is incorrect as well. (C) \(\lim_{n\to\infty} \sec^{2n}(k!\pi b) = \lim_{n\to\infty} \sec^{2n}(k! \pi (1)) = \lim_{n\to\infty} \sec^{2n}(k!\pi)\). Since k≥2, \(k!\pi\) is always a multiple of 2π, and therefore, \(\sec^{2n}(k!\pi)=1\). The limit does equal 1, so this option is correct. (D) \(\lim_{n\to\infty} \sec^{2n}(k!\pi a) = \lim_{n\to\infty} \sec^{2n}(k! \pi (1)) = \lim_{n\to\infty} \sec^{2n}(k!\pi)\). Again, since k≥2, \(k!\pi\) is always a multiple of 2π, and therefore, \(\sec^{2n}(k!\pi)=1\). The limit does equal 1, so this option is correct as well. Thus, the correct options are (C) and (D).