Question

For the limit lim x → 2 (x3 − 4x + 9) = 9 illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1

Answer 1

We can confirm the limit:\[
\lim_{x\to a} f(x) = L
\]If we can show that for any \(\epsilon\) there exists a \(\delta\) such that:\[
\forall x:\quad|x-a| < \delta \implies |f(x) - L| < \epsilon
\]

Answer 2

after i take the 9 to right side and subtract it i get 0
i take x common then what
?

Answer 3

Let's start with \(\epsilon =0.2\)

Answer 4

alright then ?

Answer 5

\[
|(x^2-4x+9)-9|<0.2
\]

Answer 6

You need to solve for \(x\) here.

Answer 7

alrights lets see

Answer 8

\[
|x^2-4x|<0.2
\]

Answer 9

\[
|a\cdot b|=|a||b|
\]

Answer 10

\[
|x||x-4|<0.2
\]

Answer 11

i make it \[x ^{3}-4x\]

Answer 12

Oh right... sorry.

Answer 13

\[
|x^3-4|<0.2\implies -0.2<x^3-4x<0.2
\]

Answer 14

hmm that seems rigth

Answer 15

For the \(\delta\) we find:\[
-\delta <x-2<\delta
\]

Answer 16

By the way \(x^3-4x=x(x^2-4)=x(x-2)(x+2)\)

Answer 17

yes i was thinkiong of that . but couldnt solve after it

Answer 18

What we basically have is: \[
\delta = \frac {0.2}{x(x+2)}
\]

Answer 19

If we let \(x=2\) at this point, then: \[
\delta = \frac{0.2}{2(2+2)}=\frac{0.2}{8}
\]

Answer 20

so now to slove for delta

Answer 21

y we taking x as 2?

Answer 22

Its \[
\lim_{x\to 2}
\]

Answer 23

my bad !

Answer 24

Though we should definitely double check our answer now.

Answer 25

i have lost all hopes with this question

Answer 26

alright let me ask someone

Answer 27

Okay sorry, we did something wrong.

Answer 28

no one to confirm bro . but if we solving for 0.1 we subsitute 0.2 right?

Answer 29

did we?

Answer 30

Yeah, I shouldn't have done \(x=2\).

Answer 31

but its tending toward 2 right?

Answer 32

Yeah, but at \(x=2\) we underestimate the epsilon.

Answer 33

Okay, I got another idea!

Answer 34

hmmm . my friend got the first one as 0.01

Answer 35

That is not the largest possible \(\delta\) though.

Answer 36

hmmm shes gonna tell the method then maybe it could help us

Answer 37

My next solution is to try letting \(x=2+\delta\)

Answer 38

\[
(2+\delta)^3-4(2+\delta)-9=9+0.2
\]

Answer 39

\[6\delta ^{2} + 8\delta+\delta ^{3} - 9\]

Answer 40

= 9+0.2

Answer 41

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%E2%88%929%3D9%2B0.2

Answer 42

that is correct if we solve it

Answer 43

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%E2%88%929%3D9%2B0.2

Answer 44

now lets see what my friend says

Answer 45

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%E2%88%929%3D9-0.2

Answer 46

\[

\delta<1.13219\\
\delta<1.11637

\]

Answer 47

so its 1.1

Answer 48

They both must be true, so that means \(1.11637\) is the largest.

Answer 49

hmmm i see
and for 0.1 we just add that pellet?

Answer 50

to the right side*

Answer 51

Oh no, I typed it in wrong before!

Answer 52

damn :P

Answer 53

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%2B9%3D9%2B0.2

Answer 54

now we got 3 solutions

Answer 55

Yeah, you always take the smallest one.

Answer 56

I mean the one that is closes to 0.

Answer 57

http://www.wolframalpha.com/input/?i=%282%2B0.024%29%5E3%E2%88%924%282%2B0.024%29%2B9%3D9%2Bepsilon

Look at how close \(|\delta| =0.024\) is!

Answer 58

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%2B9%3D9%2B0.2

Answer 59

Yeah, I'd go with \[
\delta \approx \pm 0.025485
\]

Answer 60

sweet

Answer 61

It's the lowest of all 6 deltas.

Answer 62

hmm true

Answer 63

u helped a lottt broo

Answer 64

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%2B9%3D9-0.1

http://www.wolframalpha.com/input/?i=%282%2B%CE%B4%29%5E3%E2%88%924%282%2B%CE%B4%29%2B9%3D9%2B0.1

Answer 65

When epsilon is 0.1 the lowest delta we have is: 0.0123847

Answer 66

yep saw that

Answer 67

Those are pretty accurate.

Answer 68

I feel bad using wolfram for this, but cubic roots is NOT a trivial thing.

Answer 69

The cubic formula looks like a monster that ate ten quadratic formulas.

Answer 70

hahhaha thats one way to put it:P

Answer 71

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Answer 72

See for yourself.

Answer 73

its not hard bt this question is the first one i encountered trouble with

Answer 74

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