💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Get the answer to your homework problem.

Try Numerade Free for 7 Days

Like

Report

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

$ \displaystyle \lim_{x \to 0} x^2 = 0 $

$$\begin{aligned}&\text { Given } e>0, \text { we need } \delta>0 \text { such that if } 0<|x-0|<\delta, \text { then }\left|x^{2}-0\right|<\epsilon \Leftrightarrow x^{2}<\varepsilon \Leftrightarrow|x|<\sqrt{\varepsilon} . \text { Take } \delta=\sqrt{\epsilon}\\&\text { Then } 0<|x-0|<\delta \Rightarrow\left|x^{2}-0\right|<\varepsilon . \text { Thus, } \lim _{x \rightarrow 0} x^{2}=0 \text { by the definition of a limit. }\end{aligned}$$

02:31

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

Limits

Derivatives

Harvey Mudd College

Baylor University

Idaho State University

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

01:59

Prove the statement using …

02:17

01:50

04:47

07:16

02:33

02:48

01:54

06:18

0:00

So in this problem were asked to use the epsilon delta definition of limit to prove With the limit as X approaches zero of x squared. Kind of missy. Let's clean that up is zero. All right. So, first of all, what is this epsilon delta definition say? Well, it says that for all, absolutely greater than zero. There exists Delta greater than zero. So that if x minus A is less than delta, then over as it implies that f of X minus L is less than epsilon. So What a B0 L b zero and F of X B X squared. I'm not absolutely greater than zero. Be given. Okay? So then let's choose delta equal to the square root of epsilon. All right. So then X zero because a zero which is the absolute value of X less than delta, which is the square root of epsilon. And We have f of X zero. Right? Mm hmm. Instead of. And here, I could say that this implies it affects my zero. Because L was zero is an F of X is X squared. So, this means X squared absolute X squared. All right. Is less than the absolute value of square root of epsilon squared. Well, that's the step Salon, isn't it? Because that slows growth zero anyway. So, I don't take the square or not. Take. That's the value of it. And the square the square root that just means you get what's underneath the radical there. That's excellent. So, that means I have The limit as X goes to zero then by definition of f of X, which is x squared is zero, and there is our proof.

Numerade Educator

In mathematics, precalculus is the study of functions (as opposed to calculu…

In mathematics, a function (or map) f from a set X to a set Y is a rule whic…

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a li…

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