Problem

Prove the statement using the $ \varepsilon $, $ …

03:21

Problem 19 Easy Difficulty

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

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

Answer

$\delta=\frac{3}{4} \epsilon$

Watch More Solved Questions in Chapter 2

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Problem 5

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Problem 15

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Problem 18

Problem 19

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Problem 35

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Video Transcript

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

David M.

Oklahoma State University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

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