Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 740

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{\sqrt{\left.\left. \left.\left(5+\mathrm{x}^{5}\right)-\sqrt{(5-\mathrm{x}}^{5}\right)\right\\} / \mathrm{x}^{5}\right]}=?\right.\right.\) (a) 5 (b) 25 (c) \((\sqrt{5})^{-1}\) (d) \(4 \sqrt{5}\)

Short Answer

Expert verified
(c) \((\sqrt{5})^{-1}\)

Step by step solution

01

Rewrite and simplify the expression using difference of squares formula

The expression inside the limit is \(\frac{\sqrt{(5+x^5)} - \sqrt{(5-x^5)}}{x^5}\). We can simplify this expression by multiplying it by its conjugate, which is \(\frac{\sqrt{(5+x^5)} + \sqrt{(5-x^5)}}{\sqrt{(5+x^5)} + \sqrt{(5-x^5)}}\). The new expression becomes: \(\frac{(\sqrt{(5+x^5)} - \sqrt{(5-x^5)})(\sqrt{(5+x^5)} + \sqrt{(5-x^5)})}{x^5(\sqrt{(5+x^5)} + \sqrt{(5-x^5)})}\)
02

Simplify the numerator using the difference of squares formula

Applying the difference of squares formula, the numerator simplifies to: \((5+x^5)-(5-x^5)\) Which further simplifies to: \(2x^5\)
03

Simplify the entire expression

Now, substitute the simplified numerator back into the expression: \(\frac{2x^5}{x^5(\sqrt{(5+x^5)}+\sqrt{(5-x^5)})}\) We can cancel out the \(x^5\) terms: \(\frac{2}{\sqrt{(5+x^5)}+\sqrt{(5-x^5)}}\)
04

Evaluate the limit

Now, evaluate the limit as x approaches 0: \(\lim_{x\to 0}\frac{2}{\sqrt{(5+0^5)}+\sqrt{(5-0^5)}}\) Which simplifies to: \(\frac{2}{\sqrt{5}+\sqrt{5}} = \frac{2}{2\sqrt{5}}\) Finally, we get: \(\frac{1}{\sqrt{5}}\) The answer is (c) \((\sqrt{5})^{-1}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

\(\lim _{\mathrm{x} \rightarrow[(-\pi) / 4]}[(\operatorname{Sin} 3 \mathrm{x}-\operatorname{Cos} 3 \mathrm{x}) /(4 \mathrm{x}+\pi)]=?\) (a) \(-[3 /(2 \sqrt{2})]\) (b) \([3 /(2 \sqrt{3})]\) (c) \(-[3 /(2 \sqrt{3})]\) (d) \(+[3 /(2 \sqrt{2})]\)

\(\lim _{\theta \rightarrow 0} 2[\\{\sqrt{3} \operatorname{Sin}[(\pi / 6)+\theta]-\operatorname{Cos}[(\pi / 6)+\theta]\\}\) \(/\\{\sqrt{3} \theta(\sqrt{3} \operatorname{Cos} \theta-\operatorname{Sin} \theta)\\}]=?\) (a) \((4 / 3)\) (b) \((\sqrt{3} / 4)\) (c) \((4 / 9)\) (d) \((2 / 3)\)

The value of \(\mathrm{m}\) and \(\mathrm{n}\) for which the function \(f(x)=\mid \begin{array}{ll}{[\\{\operatorname{Sin}(m+1) x+\sin x\\} / x],} & x<0 \\ n, & x=0 \\ {\left[\left\\{\sqrt{\left. \left.\left(x+x^{2}\right)-\sqrt{x}\right\\} /\left(x^{(3 / 2)}\right)\right],}\right.\right.} & x>0\end{array}\) is continuous for \(\forall x \in R\) ? (a) \(\mathrm{m}=-(3 / 2), \mathrm{n}=(1 / 2)\) (b) \(\mathrm{m}=(1 / 2), \mathrm{n}=(3 / 2)\) (c) \(\mathrm{m}=(1 / 2), \mathrm{n}=-(3 / 2)\) (d) \(\mathrm{m}=(5 / 2), \mathrm{n}=(1 / 2)\)

\(\lim _{\mathrm{x} \rightarrow(\pi / 3)}[\\{1-\sqrt{(3) \cot x}\\} /(2 \cos x-1)]=?\) (a) \((4 / 3)\) (b) \(-(4 / 3)\) (c) \((2 / 3)\) (d) \(-(2 / 3)\)

\(\lim _{\mathrm{x} \rightarrow 0}\left(3 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+2 \mathrm{x}\right)-\left(3 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+\mathrm{x}\right)\) \(-\left(1 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+3 \mathrm{x}\right)+\left(1 / \mathrm{x}^{3}\right) \sin [\pi(1+\pi)]=?\) (a) \(\operatorname{Cos} \pi^{2}\) (b) \(-\operatorname{Cos} \pi^{2}\) (c) \(-\pi\) (d) \(\pi\)
See all solutions

Recommended explanations on English Textbooks

English Grammar

Read Explanation

Lexis and Semantics

Read Explanation

Graphology

Read Explanation

Essay Writing Skills

Read Explanation

Phonetics

Read Explanation

Language Acquisition

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.

Sign-up for free