/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 Given $l_{1}=\lim _{x \rightarro... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 86

Given $l_{1}=\lim _{x \rightarrow \frac{\pi}{4}} \cos ^{-1}\left[\sec \left(x-\frac{\pi}{4}\right)\right]$; $l_{2}=\lim _{x \rightarrow \frac{\pi}{4}} \sin ^{-1}\left[\operatorname{cosec}\left(x+\frac{\pi}{4}\right)\right]$ $l_{3}=\lim _{x \rightarrow \frac{\pi}{4}} \tan ^{-1}\left[\cot \left(x+\frac{\pi}{4}\right)\right]$ $l_{4}=\lim _{x \rightarrow \frac{\pi}{4}} \cot ^{-1}\left[\tan \left(x-\frac{\pi}{4}\right)\right]$ where \([\mathrm{x}]\) denotes greatest integer function then which of the following limits exist (A) \(l_{1}\) (B) \(l_{2}\) (D) \(l_{4}\) (C) \(l_{3}\)

Short Answer

Expert verified
a) \(l_{1}\) b) \(l_{2}\) c) \(l_{3}\) d) \(l_{4}\) Answer: All of the given limits, \(l_1, l_2, l_3,\) and \(l_4\), exist.

Step by step solution

01

Compute limit \(l_{1}\)

To compute the limit \(l_{1}\), we have $$l_{1} = \lim _{x \rightarrow \frac{\pi}{4}} \cos ^{-1}\left[\sec\left(x-\frac{\pi}{4}\right)\right].$$ Using the identities: \(\cos^{-1}(x) = \pi/2 - \sin^{-1}(x)\) and \(\sec(x) = 1/\cos(x)\), we can modify the limit as $$l_1 = \lim_{x\rightarrow\frac{\pi}{4}}\left(\frac{\pi}{2} - \sin^{-1}\left(\cos\left(x-\frac{\pi}{4}\right)\right)\right).$$ Now evaluate the limit, using the fact that \(\sin^{-1}(\sin(x)) = x\), $$l_1 = \frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \frac{\pi}{2}.$$ So, the limit \(l_{1}\) exists.
02

Compute limit \(l_{2}\)

To compute the limit \(l_{2}\), we have $$l_{2} = \lim _{x \rightarrow \frac{\pi}{4}} \sin ^{-1}\left[\operatorname{cosec}\left(x+\frac{\pi}{4}\right)\right].$$Using the identity: \(\cosec(x) = 1/\sin(x)\), we can modify the limit as $$l_2 = \lim_{x\rightarrow\frac{\pi}{4}} \sin^{-1}\left(\frac{1}{\sin\left(x +\frac{\pi}{4}\right)}\right).$$ Now evaluate the limit, using the fact that \(\sin^{-1}(\sin(x)) = x\), $$l_2 = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.$$ So, the limit \(l_{2}\) exists.
03

Compute limit \(l_{3}\)

To compute the limit \(l_{3}\), we have $$l_{3}=\lim _{x \rightarrow \frac{\pi}{4}} \tan ^{-1}\left[\cot \left(x+\frac{\pi}{4}\right)\right].$$ Using the identities: \(\tan^{-1}(x) = \pi/2 - \cot^{-1}(x)\) and \(\cot(x) = 1/\tan(x)\), we can modify the limit as $$l_3 = \lim_{x\rightarrow\frac{\pi}{4}}\left(\frac{\pi}{2} - \cot^{-1}\left(\tan\left(x+\frac{\pi}{4}\right)\right)\right).$$ Now evaluate the limit, using the fact that \(\cot^{-1}(\cot(x)) = x\), $$l_3 = \frac{\pi}{2} - \left(\frac{\pi}{4} + \frac{\pi}{4}\right) = 0.$$ So, the limit \(l_{3}\) exists.
04

Compute limit \(l_{4}\)

To compute the limit \(l_{4}\), we have $$l_{4}=\lim _{x \rightarrow \frac{\pi}{4}} \cot ^{-1}\left[\tan \left(x-\frac{\pi}{4}\right)\right].$$ Using the identities: \(\cot^{-1}(x) = \pi/2 - \tan^{-1}(x)\) and \(\tan(x) = 1/\cot(x)\), we can modify the limit as $$l_4 = \lim_{x\rightarrow\frac{\pi}{4}}\left(\frac{\pi}{2} - \tan^{-1}\left(\cot\left(x-\frac{\pi}{4}\right)\right)\right).$$ Now evaluate the limit, using the fact that \(\tan^{-1}(\tan(x)) = x\), $$l_4 = \frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{\pi}{4}\right) = \frac{\pi}{2}.$$ So, the limit \(l_{4}\) exists. Now, all the limits \(l_1, l_2, l_3,\) and \(l_4\) exist. Therefore, the correct answer is all options (A), (B), (C), and (D).

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Most popular questions from this chapter

Assertion \((A):\) A circle \(C_{1}\) is inscribed in an equilateral triangle \(\mathrm{ABC}\) with side length \(2 .\) Then circle \(\mathrm{C}_{2}\) is inscribed tangent to BC, CA and circle \(\mathrm{C}_{1}\). An infinite sequence of such circles is constructed, each tangent to \(\mathrm{BC}, \mathrm{CA}\) and the previous circle. The sum of areas of all the infinitely many circles is \(\frac{5 \pi}{8}\). Reason ( \(\mathbf{R}\) ) : Radius of \(\mathrm{C}_{1}\) is \(\frac{1}{\sqrt{3}}\), that of \(\mathrm{C}_{2}\) is \(\frac{1}{3 \sqrt{3}}\) and radius of the remaining circle each shrink by a factor \(\frac{1}{3}\).

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 \(l_{1}\).

$\lim _{x \rightarrow \infty} x^{2}\left[\tan ^{-1} \frac{2 x^{2}+1}{x^{2}+2}-\tan ^{-1} 2\right]$ is (A) \(\frac{3}{5}\) (B) \(-\frac{3}{5}\) (C) \(\frac{5}{3}\) (D) \(-\frac{5}{3}\)

Column - I (A) If $\lim _{x \rightarrow \infty}\left(\sqrt{\left(x^{2}-x-1\right)}-a x-b\right)=0\(, where \)a>0$, then there exists atleast one a and \(b\) for which point (a, \(2 b\) ) lies on the line (B) If $\lim _{x \rightarrow 0} \frac{\left(1+a^{3}\right)+8 e^{1 / x}}{1+\left(1-b^{3}\right) e^{1 / x}}=2$, then there exists atleast one \(a\) and \(b\) for which point \(\left(a, b^{3}\right)\) lies on the line (C) If $\lim _{\mathrm{x} \rightarrow \infty}\left(\sqrt{\left(\mathrm{x}^{4}-\mathrm{x}^{2}+1\right)}-\mathrm{ax}^{2}-\mathrm{b}\right)=0$, then there exists atleast one a and \(\mathrm{b}\) for which point \((\mathrm{a},-2 \mathrm{~b})\) lies on the line (D) If \(\lim _{x \rightarrow-a} \frac{x^{7}+a^{7}}{x+a}=7\), where \(a<0\), then there exists atleast one a for which point \((a, 2)\) lies on the line. Column - II (P) \(\mathrm{y}=-3\) (Q) \(3 x-2 y-5=0\) (R) \(15 x-2 y-13=0\) (S) \(\mathrm{y}=2\)

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