Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Question

Answered step-by-step

Find the limit, if it exists. If the limit does not exist, explain why.

$ \displaystyle \lim_{x \to -2}\frac{2 - |x|}{2 + x} $

Video Answer

Solved by verified expert

This problem has been solved!

Try Numerade free for 7 days

Like

Report

Official textbook answer

Video by Daniel Jaimes

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Limits

Derivatives

University of Nottingham

Idaho State University

Boston College

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

00:36

Find the limit if it exist…

02:24

Find the limit, if it exis…

00:33

Find the limit (if it exis…

00:23

00:26

00:58

01:25

Evaluate the limit or expl…

01:22

00:31

02:03

This is problem number forty four of the Stuart Calculus. Eighth edition, section two point three. Final amid. If it exists, it's the limit does not exist. Explain why Here we have a limited six bridges. Negative too. Of the function to minus work. The absolute value backs divide where the quantity two plus sex. Okay, since this is a complete limit, not just a limit from the left or the right, we wantto determine first. Whether approach in a tutu makes this an indeterminate form or an undefined value. Two minus the absolutely of negative too is to minus two two plus native who? Zero. So we have a zero division here, divide by zero. Add some *** too. So we know that the value or the function is on to find a native to so we were unable to determine that it exists are at this point. Um, but we can still prove that ism it exists as long as the limited left negative two equals limit from the right. I'm negative too. On that the equal each other. So what we want to do is we want to first deal with this absolute value function, and we want to rewrite this Lim. I'm using what we know about the absolute value function of X. We know that when X is greater than zero or equal to zero, it's equal the function X. However, when X is negative, he also value is meant to change the sign and so dysfunctions represented by a negative X and for our interests since the limit of expertise and negative too. This is on ly in this region. So we will not be considering the absolute value function too. Take this form because we are only concerned with the X approaches negative to region. So we will substitute dysfunction with negative X instead. So rewriting this limit as experts needed to two minus native X or plus X divided by two plus X Now notice that I didn't write experts in ecstasy from the left or expert unit two from the right because both of those give you the same results right since we're pushing native to which is a value in this region. So the function looks the same, whether it's from the left or the right. As we can see, we can cancel these too, and this simplifies to one And if we take the Limited's experts think, too, of this constant value of one. Her answer for the limit is one again keeping in mind that this is true for both. Pushing is equal to negative to pushing that from the left and from the right. And since those Walter True and exist and they both equal one, then this limit from the beginning also equals one, and that is our final answer.

View More Answers From This Book

Find Another Textbook