Question

A positive number and the limit L of a function f at a are given. Find a number δ such that |f (x) − L| <  if 0 < |x − a| < δ.
https://i.gyazo.com/1ee47f23fea66c55f7287e49aaec8394.png

Answer 1

\[|x^3-8|<.001\] solve for \(|x-2|<\delta\)
hint, factor the difference of two cubes

Answer 2

what's delta supposed to be?

Answer 3

that is what you have to find

Answer 4

2 something?

Answer 5

\[|x^3-8|=|(x-2)(x^2+2x+4)|=|x-2||x^2+2x+4|<.001\]

Answer 6

12?

Answer 7

find a bound for \(|x^2+2x+4|\)

Answer 8

11.999 and 12.001?

Answer 9

for example if \(x<1\) then \(x^2+2x+4<7\)

Answer 10

\(\delta\) will be a small number, not a large one `

Answer 11

where did the 7 come from?

Answer 12

i picked \(|x<1\)

Answer 13

wait so it's zero?

Answer 14

no \(\delta\) is positive

Answer 15

how do i solve that?

Answer 16

i plugged in 2 but that didn't work

Answer 17

you want \[|x-2||x^2+2x+4|<.001\]and you have control over \(|x-2|\)

Answer 18

if say \(x<1\) then \(|x^2+2x+4|<7\) i think
that means you have to make sure that \[|x-2|\times 7<.001\]

Answer 19

divide by 7 should do it

Answer 20

wait that is wrong sorry

Answer 21

i dont understand what you're asking me to do?

Answer 22

ok i made a mistake, but lets go slow and we can correct it

Answer 23

you want to find how small to make \(|x-2|\) so that \(|x^3-8|<.001\)

Answer 24

is that part clear?
your answer has to be \(|x-2|<\text{some expression with .001}\)

Answer 25

i think i'm just going to drop this class

Answer 26

lol it is not really that bad, and it gets easier
is this calc 1?

Answer 27

yes

Answer 28

ok this kind of problem is different from the kind you are probably used to, and also different from most other calc questions
no need to drop because of this
the hardest part is understanding what it means and what you need to do
after this, the problems get simpler

Answer 29

in fact, many classes either skip this or gloss over it real quick

Answer 30

ooh okay. I'm going to try to read my textbook. Hopefully this type of question isn't on my quiz tomorrow

Answer 31

hope not, but we can figure it out anyways if you like
will take a few minutes, up to you

Answer 32

yes but i'm not sure how to solve it

Answer 33

ok so lets start by understanding what the question is really asking
here are some things that should be clear
if \(x=2\) then \(x^3=8\) right?

Answer 34

yes

Answer 35

oops i meant "if \(x=2\) then \(x^3-8=0\) yes?

Answer 36

yes

Answer 37

what you are going to show is pretty obvious (not obvious to show, just an obvious fact) that if \(x\) is close to \(2\) then \(x^3\) is close to \(8\)

Answer 38

they way you say \(x^3\)is close to \(8\) in math is \(|x^3-8|\) is close to 0

Answer 39

the absolute value because it could be a little bigger than 8 or a little smaller

Answer 40

so far so good?

Answer 41

yes

Answer 42

so what you are trying to show is the obvious fact that if \(x\) is close to \(2\) then \(x^3\) is close to \(8\) which, when we write that in math, is \(|x^3-8|\) is small if \(|x-2|\) is small
the first statement says \(x^3\) is close to \(8\) and the second says \(x\) is close to \(2\)

Answer 43

how close do you have to make \(x^3\) to \(8\)? in this question they are asking you to make it within \(.001\) of \(8\) translates as \[|x^3-8|<.001\]

Answer 44

and you have to figure out how close to make \(x\) to \(2\) in order for that to be true, i.e. you have to find some (probably very very small) number to insure that if \(|x-2|<\text{that number}\) then \(|x^3-8|<.001\)
you are looking for "that number"

Answer 45

any number?

Answer 46

no, that number will depend in a couple things, but on thing it depends on is \(.001\)

Answer 47

that is what you are trying to find, that is all the work for this question

Answer 48

1.999 and 2.001?

Answer 49

well IF \(x \) is between \(1.99\) and \(2.001\) then for sure \(|x-2|<.001\) but that not might be good enough to make \(|x^3-8|<.001\) you might have to take something smaller

Answer 50

now it it clear what we have to find? some number , which they are calling \(\delta\)
so that if \(|x-2|<\delta\) then \(|x^3-8|<.001\)

Answer 51

we are searching for the right \(\delta\)
that is all the work, besides understanding the question

Answer 52

x^3<8.001?

Answer 53

@satellite73 can you help me ???/

Answer 54

yes, if \[|x^3-8|<.001\] then \[7.999<x^3<8.001\] but we still have to find \(\delta\)

Answer 55

that is all the work, finding the right \(\delta\)
think we can handle it?

Answer 56

do i have to find the cube of 8.001

Answer 57

no, not at all
we have to do a bit of algebra

Answer 58

or we could, which is not what your math teacher wants, do it by trial and error

Answer 59

could i just put cube root (8.001)?

Answer 60

we can try that and see if it works

Answer 61

http://www.wolframalpha.com/input/?i=cuberoot%288.001%29

Answer 62

so for sure \(x<2.00007\) would work

Answer 63

now lets try \(\sqrt[3]{7.999}\)

Answer 64

1.999916663

Answer 65

http://www.wolframalpha.com/input/?i=cuberoot%287.999%29

Answer 66

ok, this is not what your math teacher wants you to do, but it will work
we can say \[1.9993<x<2.0007\]

Answer 67

which means in other words that \[|x-2|<.00007\]

Answer 68

i had the wrong number of 9's there \[1.99993<x<2.00007\]

Answer 69

which does mean \[|x-2|<.00007\]

Answer 70

i think this is as good an explanation and method as any
it is not as abstract as it could be, but at least we understand what is going on
get an upper and lower bound for \(x\) and sandwich it between them

Answer 71

thank you!

Answer 72

yw, hope it is a bit more clear than it was at the start