Question

By the method of delta-epsilon prove

limx->5 ((1/2)x+6)= 8 and a1/2

Answer 1

ugh.... epsilons and deltas.....

Answer 2

f(x)-L < e

|x-c|<d

Answer 3

at least its a line and not a curve

Answer 4

i know all of that sorry let me tell you where am having a problem

Answer 5

i know all of that sorry let me tell you where am having a problem

Answer 6

ok i simplified it to get 1/2 (x-2) < e

so how do i find my delta from there?

Answer 7

|x-5| < d right? id have to work it out meself to recall how to do it lol

Answer 8

yes

Answer 9

yes

Answer 10

its |x-5|<d

Answer 11

a1/2 ?? whats that mean?

-e < (1/2)x+6) - 8 < e

Answer 12

8 and a half

Answer 13

-e < (1/2)x-2 < e

2 -e < (1/2)x < 2+e

2(2 -e) < x < 2(2+e)

Answer 14

0 < x-5 <d

5 < x < 5+d

Answer 15

5+d < 2(2+e) i got know idea if this is working lol

Answer 16

arent we suppose to be converting x-2 to x-5?..no?

Answer 17

usually you work these problems backwards.

Answer 18

this one is tricky...

Answer 19

start with |
\[|.5x+6-8.5|<\epsilon\]

Answer 20

yes

Answer 21

\[-\epsilon < .5x-2.5< \epsilon\]

Answer 22

ohh....8' 1/2 ....

Answer 23

\[2.5-\epsilon<.5x<2.5+\epsilon\]

Answer 24

\[5-2\epsilon<x<5+2\epsilon\]

Answer 25

looks familiar so far lol

Answer 26

can u do it in terms of epsilon and delta?

Answer 27

can u do it in terms of epsilon and delta?

Answer 28

how am i doing so far?

Answer 29

good

Answer 30

well, other than having the right numbers; looks the same as mine lol

Answer 31

oh my delta will depend on epsilon. i am just trying to figure out what the delta needs tgo be

Answer 32

kk

Answer 33

the game is played like this: i say the limit is L. you say ok, make it within epsilon of L. i say ok use this delta which obviously depends on epsilon. that is delta will be written in terms of epsilon

Answer 34

ok but in class we normally would deleta write where u wrote your epsilon but its problem a different example

Answer 35

wait

Answer 36

how'd you get the .5 to 5?

Answer 37

i am working the problem backwards. then the proof is to work it forwards. you say for example pick delta such that \[5-2\epsilon < \delta< 5+2\epsilon\]

Answer 38

the 5?

Answer 39

we had
\[2.5- \epsilon<.5x<2.5+\epsilon\]

Answer 40

to get x by itself i doubled everything

Answer 41

ok so now we have
\[5-2\epsilon <x < 5+2\epsilon\]

Answer 42

this says the same thing as \[|x-5|<2\epsilon\]

Answer 43

so the delta i pick is
\[2\epsilon\]

Answer 44

oh ok sorry i had to move from the pc a sec

Answer 45

because that is what we wanted. we wanted to say that we could find a delta such that if \[|x-5|<\delta\] then \[|.5x-6+8.5|<\epsilon\]

Answer 46

don't forget what the goal is. you want to show that
\[lim_{x->5} (.5x+6)=8\]

Answer 47

8 1/2

Answer 48

meaning that given any epsilon > 0 y ou can find a delta such that if
\[|x-5|<\delta\] then \[|.5x+6-8.5|<\epsilon\]

Answer 49

ok truthfully i dont understand what u did

Answer 50

i say pick \[\delta = 2 \epsilon\] i win by running the algebra backwards

Answer 51

the way you have to do these problem is work backwards. you have to find the delta, which depends on epsilon. i.e. \[\delta = delta(\epsilon)\]

Answer 52

delta will be an expression in epsilon. so you start with the thing that you want, namely
\[|.5x+6-8.5|<\epsilon\] do some algebra, and see what delta you get.

Answer 53

i got
\[\delta = 2\epsilon\]

Answer 54

ok

Answer 55

thank you!

Answer 56

run the algebra forward and see that it works. just take the steps i wrote and write them backwards. in other words start with
\[|x-5|<2\delta\] and see that you get
\[|2.x+6-8.5|<\epsilon\]

Answer 57

typo

Answer 58

i meant start with \[|x-5|<2\epsilon\]

Answer 59

i will write it if you like

Answer 60

i dont undestand what your asking me to do

Answer 61

let me write it and maybe it will make sense. hold on and i will write it all out.

Answer 62

okat

Answer 63

You want to show that given any \[\epsilon>0\] you can find a \[\delta\] such that if

\[|x-5|<\delta\] then \[|.5x+6-8.5|<\epsilon\]

let
\[\delta=2\epsilon\]
then
\[|x-5|<2\epsilon\]
means
\[-2\epsilon<x-5<\epsilon\]
giving

\[5-2\epsilon < x < 5+2\epsilon\] gives
\[2.5-\epsilon < .5x < 2.5 +\epsilon\] so
\[-\epsilon < .5x-2.5<\epsilon\] or \[|.5x-2.5|<\epsilon\]

Answer 64

sorry it took so long to write. but i never would have found delta = 2 epsilon without working the problem backwards

Answer 65

hope this helps

Answer 66

wow THANK YOU SO MUCH

Answer 67

welcome.

Answer 68

ok like thank you so much i get it but i never would have figured that out on my own thanks a billion seriously