Question

use abslon-delta to show that f(x)=x^2 AT x dote =2

Answer 1

Abslon!

Answer 2

what is \(x\) approaching?

Answer 3

yes abslon

Answer 4

epsilon :)

Answer 5

what is dote?

Answer 6

xo

Answer 7

what is xo ?

Answer 8

\[\lim_{x\rightarrow?}x^2\]

Answer 9

its continuous property

Answer 10

when we use epsilon - delta and say x^2-2 <EPSLION

Answer 11

Does x approach \(2\)? Does x approach \(\sqrt 2\)?

Answer 12

the first on x approach 2

Answer 13

f(x)=x^2 at x0 =2

Answer 14

Well as \(x\to 2\) then \(x^2\to 4\). Right?

Answer 15

\[
|x-2|<\delta \implies |x^2-4|<\epsilon
\]

Answer 16

\[
|x-2|<\delta \implies |x-2||x+2|<\epsilon
\]

Answer 17

I have to say first and prove /x^2-2/epsilon

Answer 18

< epsilon

Answer 19

Okay sorry, but you obviously are confused..

Answer 20

\[
\lim_{x\to a}f(x)=L
\]You need to decide what \(a\) and \(L\) are. Let me assure you they both can't be \(2\) because \(2^2\neq 2\)

Answer 21

yes sorry I know because its hard to write the symbol

Answer 22

yes I think I have to prove that f(x) is continues if f(x)=x^2 at x0= 2

Answer 23

to denote \(x_0\) write x_0

Answer 24

ok do you need to prove something is continuous or do you need to prove the limit?

Answer 25

okay sorry about yhat

Answer 26

no need to be sorry:)

Answer 27

I need to prove it continues

Answer 28

so you need to prove x^2 is continuous at x = 2?

Answer 29

RIGHT

Answer 30

So show \(f(2) = L\) and then show \(\lim_{x\to 2}f(x) = L\)

Answer 31

No abslons needed.

Answer 32

no in the question say use epsilon - delta

Answer 33

Okay so first let's find out \(L\). Can you do that?