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Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to 1}\frac{2 + 4x}{3} = 2 $

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03:23

Daniel Jaimes

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

Limits

Derivatives

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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.

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Prove the statement using …

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$19-32$ Prove the statemen…

So in this problem, were asked to prove That the limit as X approaches one of two plus four, X over three equals two. Using the epsilon delta definition. Well, the epsilon delta definition says that if x minus one, so that's in delta. And that implies that f of x minus L is less than epsilon. Where L is the limit. All right. So, We're gonna let ε greater than zero be given then. We're gonna choose delta Equal to 3/4 times epsilon. Mhm. Okay. So then we have we'll see this was the x minus a. Sorry? So a is one F of X is two plus four X over three, and L is too All right. So then I have observations X minus one. Less than delta implies. What? That F of x minus L. Well, that's the value of two plus four. X over three minus two. Okay. And this is the absolute value of two plus four. X minus six. All over three. Well, So then this is four X -4. Yeah, civilian over three. Which is absolutely of 4/3, which that's already positive. So we can just factor it out straight like it is Comes up the value of X -1. Right, Okay. But X -1 is less than Delta. So this is less than 4/3 Delta. And Delta was three or 4 ε. This is 4/3 times 3/4 epsilon. And these cancel. So this is epsilon. So I mean fx minus l is less than epsilon. So therefore we have no the limit As X approaches one of two plus four, X over three equals two. Bye, epsilon delta definition. Yeah. And

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