/*! 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 12 The value of $\lim _{x \rightarr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 12

The value of $\lim _{x \rightarrow 2}\left(\left(\frac{x^{3}-4 x}{x^{3}-8}\right)^{-1}-\left(\frac{x+\sqrt{2 x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right)$ is (A) \(1 / 2\) (B) 2 (C) 1 (D) None of these

Short Answer

Expert verified
After analyzing and solving the given limit expression step by step, we found out that the limit is equal to the inverse of \(3\sqrt{2}\), which did not match any of the given options. Therefore, the answer is (D) None of these.

Step by step solution

01

Simplify the first sub-expression

To simplify the first sub-expression inside the limit, we will factor out the numerator and denominator: \(\left(\frac{x^{3}-4x}{x^{3}-8}\right)^{-1} = \left(\frac{x(x^2-4)}{x^3-8}\right)^{-1} = \left(\frac{x(x+2)(x-2)}{(x-2)(x^2+2x+4)}\right)^{-1}\) Now, we can cancel out the common factors from the numerator and denominator: \(\left(\frac{x(x+2)(x-2)}{(x-2)(x^2+2x+4)}\right)^{-1} = \left(\frac{x(x+2)}{x^2+2x+4}\right)^{-1}\)
02

Simplify the second sub-expression

Now we will simplify the second sub-expression inside the limit by finding the least common denominator (LCD) to combine the two fractions: \(\left(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1} = \left(\frac{(x+\sqrt{2x})(\sqrt{x}+\sqrt{2}) - \sqrt{2}(x-2)}{(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\)
03

Find common denominator and simplify

Now, we have the simplified sub-expressions, and we can find the LCD to combine them: \(\lim_{x \rightarrow 2}\left(\left(\frac{x(x+2)}{x^2+2x+4}\right)^{-1}-\left(\frac{(x+\sqrt{2x})(\sqrt{x}+\sqrt{2}) - \sqrt{2}(x-2)}{(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\right)\) The least common denominator (LCD) is \((x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2})\). Therefore, we will multiply the numerators and denominators by the appropriate factors to combine the fractions: \(\lim_{x \rightarrow 2} \left(\frac{(x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2}) - x(x+2)(x-2)(\sqrt{x}+\sqrt{2})}{(x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\) Now simplify the expression inside the limit: \(\lim_{x \rightarrow 2} \left(\frac{(x^2+2x+4)(\sqrt{x}+\sqrt{2})(x-2)}{(x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\) Cancel out the common factors: \(\lim_{x \rightarrow 2} \left(\frac{x^2+2x+4}{\sqrt{x}+\sqrt{2}}\right)^{-1}\) Now, we can evaluate the limit by substituting x = 2: \(\left(\frac{2^2+2(2)+4}{\sqrt{2}+\sqrt{2}}\right)^{-1} = \left(\frac{12}{2\sqrt{2}}\right)^{-1} = \left(\frac{6}{\sqrt{2}}\right)^{-1}\) Now, multiply the top and bottom with \(\sqrt{2}\): \(\left(\frac{6\sqrt{2}}{2}\right)^{-1} = \left(3\sqrt{2}\right)^{-1}\) So, the value of the given limit is the inverse of \(3\sqrt{2}\): \(\lim_{x \rightarrow 2}\left(\left(\frac{x^{3}-4x}{x^{3}-8}\right)^{-1}-\left(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right) = \left(3\sqrt{2}\right)^{-1}\) Thus, the answer is (D) None of these.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that a and \(\mathrm{b}\) are real positive numbers then the value of \(\lim _{t \rightarrow 0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1 / t}\) has the value equals to (A) $\frac{\mathrm{a} \ln \mathrm{b}-\mathrm{b} \ln \mathrm{a}}{\mathrm{b}-\mathrm{a}}$ (B) $\frac{\mathrm{b} \ln \mathrm{b}-\mathrm{a} \ln \mathrm{a}}{\mathrm{b}-\mathrm{a}}$ (C) \(\mathrm{b} \ln \mathrm{b}-\mathrm{a} \ln \mathrm{a}\) (D) \(\left(\frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}\)

If $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{2 \mathrm{x}}{\pi} \tan ^{-1} \mathrm{nx}\(, then value of \)\lim _{\mathrm{x} \rightarrow 0}[\mathrm{f}(\mathrm{x})-1]$ is, where [.] represents greatest integer function (A) 0 (B) \(-1\) (C) 1 (D) does not exist

The value of $\lim _{n \rightarrow \infty} \frac{1^{6}+2^{6}+3^{6} \ldots \ldots . n^{6}}{\left(1^{2}+2^{2}+3^{2} \ldots \ldots n^{2}\right)\left(1^{3}+2^{3}+3^{3}+\ldots \ldots . n^{3}\right)}$ (A) \(\frac{14}{7}\) (B) \(\frac{21}{8}\) (C) \(\frac{132}{17}\) (D) \(\frac{12}{7}\)

\(\lim _{x \rightarrow 1} \frac{\tan (x-1) \cdot \log _{e} x^{x-1}}{|x-1|^{3}}\) is equal to (A) 1 (B) \(-1\) (C) 3 (D) None of these

Which of the following limits exist? (where [.] indicates greatest integer function all throughout) (A) \(\lim _{x \rightarrow 1} \frac{\sin [x]}{[x]}\) (B) $\lim _{n \rightarrow \infty}\left(\frac{\mathrm{e}^{\mathrm{n}}}{\pi}\right)^{1 / \mathrm{n}}$ (C) \(\lim _{x \rightarrow 1}\left[\sin \left(\sin ^{-1} x\right)\right]\) (D) \(\lim _{x \rightarrow \pi / 2}\left[\sin ^{-1}(\sin x)\right]\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.