Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Exercises \(7 - 14 ,\) determine the limit by substitution. Support graphically. $$\lim _ { x \rightarrow - 2 } ( x - 6 ) ^ { 2 / 3 }$$

Short Answer

Expert verified
Therefore, the solution to \(\lim _ { x \rightarrow - 2 } (x - 6) ^ {2 / 3}\) is \(4\).

Step by step solution

01

Substitute x in the function

According to the limit's substitution principle, first, we need to substitute \(x = -2\) into the function \((x - 6)^{2/3}\). So our function will change into \((-2 - 6)^{2/3}\).
02

Solving the Expression

After substituting \(x=-2\) into the function, let's solve the equation \((-2 - 6)^{2/3}\). First, calculate the result of \(-2 - 6\), which equals \(-8\). Now the equation becomes \((-8)^{2/3}\.
03

Calculating the Cube Root

Next, let's simplify the equation further. The cube root of -8 is -2. So, \((-8)^{2/3}\) equals to \((-2)^2\), which is \(4\). Therefore, the limit of the function \((x - 6)^{2/3}\) as \(x\) approaches \(-2\) is \(4\).
04

Graphical Representation

When graphing the function \((x - 6)^{2/3}\), we will observe that as \(x\) approaches \(-2\), the function approaches \(4\). This graphical representation confirms the limit found algebraically.

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 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

See all solutions

Recommended explanations on Math Textbooks

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