Question

Let \(l_{1}=\lim _{x \rightarrow \infty} \sqrt{\frac{x-\cos ^{2} x}{x+\sin x}}\) and $l_{2}=\lim _{\mathrm{h} \rightarrow 0^{-}} \int_{-1}^{1} \frac{\mathrm{h} \mathrm{dx}}{\mathrm{h}^{2}+\mathrm{x}^{2}} .$ Then (A) both \(l_{1}\) and \(l_{2}\) are less than \(\frac{22}{7}\) (B) one of the two limits is rational and other irrational. (C) \(l_{2}>l_{1}\) (D) \(l_{2}\) is greater than 3 times of \(l_{1}\).

Short answer

In this problem, we are given two limits, \(l_1\) and \(l_2\), and are asked to determine which of the provided statements are true based on their values. We compute \(l_1\) by taking the limit as \(x\) approaches infinity of a given expression, resulting in a value of 1. We compute \(l_2\) by evaluating an improper integral with a limit as \(h\) approaches 0 from the negative side, resulting in a value of \(\pi\). Based on these values, we find that statements (A), (B), and (C) are true while statement (D) is false.

Step-by-step solution

Calculate \(l_1\)

To find the value of \(l_1\), we need to evaluate the given limit: \(l_{1} = \lim_{x \to \infty} \sqrt{\frac{x-\cos^{2} x}{x+\sin x}}\) First, let's rewrite the limit as a fraction of two square roots: \(l_{1} = \lim_{x \to \infty}\frac{\sqrt{x-\cos^{2} x}}{\sqrt{x+\sin x}}\) One way to solve this limit is to divide both numerator and denominator by \(x\). This simplifies the expression: \(l_{1} = \lim_{x \to \infty}\frac{\sqrt{1-\frac{\cos^{2} x}{x}}}{\sqrt{1+\frac{\sin x}{x}}}\) Since \(\lim_{x \to \infty}\frac{\cos^{2} x}{x} = 0\) and \(\lim_{x \to \infty}\frac{\sin x}{x} = 0\), we can write: \(l_{1} = \frac{\sqrt{1-0}}{\sqrt{1+0}}\) Therefore, \(l_1 = 1\).

Calculate \(l_2\)

To calculate \(l_2\), we need to evaluate the improper integral: \(l_{2} = \lim_{h \to 0^-} \int_{-1}^{1} \frac{h \, dx}{h^{2}+x^{2}}\) To do this, we first evaluate the integral and then calculate the limit as \(h\to 0^-\). Let's evaluate the integral: \(\int_{-1}^{1} \frac{h \, dx}{h^{2}+x^{2}}\) = \(\left[Arctan({\frac{x}{h}})\right]_{-1}^{1}\) Now, let's apply the limit: \(l_{2} = \lim_{h\to 0^-} (Arctan({\frac{1}{h}}) - Arctan({\frac{-1}{h}}))\) Since we know that \(\lim_{x\to\infty}Arctan(x) = \frac{\pi}{2}\) and \(\lim_{x\to-\infty}Arctan(x) = -\frac{\pi}{2}\), we can rewrite the limit as: \(l_{2} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi\) Therefore, \(l_2 = \pi\).

Comparing \(l_1\) and \(l_2\)

Now that we have calculated both \(l_1 = 1\) and \(l_2 = \pi\), we can verify the given statements: (A) both \(l_{1}\) and \(l_{2}\) are less than \(\frac{22}{7}\): Since \(1 < \frac{22}{7}\) and \(\pi < \frac{22}{7}\) (approximately), this statement is true. (B) one of the two limits is rational and the other irrational: \(l_1 = 1\) (rational) and \(l_2 = \pi\) (irrational), so this statement is also true. (C) \(l_{2}>l_{1}\): Since \(\pi > 1\), this statement is true. (D) \(l_{2}\) is greater than 3 times of \(l_{1}\): Since \(\pi > 3 (\approx 3.14),\) this statement is false. The correct options are (A), (B), and (C).