By the method of delta-epsilon prove
limx->5 ((1/2)x+6)= 8 and a1/2

- anonymous

By the method of delta-epsilon prove
limx->5 ((1/2)x+6)= 8 and a1/2

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- amistre64

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

- amistre64

f(x)-L < e
|x-c|

- amistre64

at least its a line and not a curve

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## More answers

- anonymous

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

- anonymous

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

- anonymous

ok i simplified it to get 1/2 (x-2) < e
so how do i find my delta from there?

- amistre64

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

- anonymous

yes

- anonymous

yes

- anonymous

its |x-5|

- amistre64

a1/2 ?? whats that mean?
-e < (1/2)x+6) - 8 < e

- anonymous

8 and a half

- amistre64

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

- amistre64

0 < x-5

- amistre64

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

- anonymous

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

- anonymous

usually you work these problems backwards.

- anonymous

this one is tricky...

- anonymous

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

- anonymous

yes

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

\[5-2\epsilon

- amistre64

looks familiar so far lol

- anonymous

can u do it in terms of epsilon and delta?

- anonymous

can u do it in terms of epsilon and delta?

- anonymous

how am i doing so far?

- anonymous

good

- amistre64

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

- anonymous

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

- anonymous

kk

- anonymous

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

- anonymous

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

- anonymous

wait

- anonymous

how'd you get the .5 to 5?

- anonymous

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

- anonymous

the 5?

- anonymous

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

- anonymous

to get x by itself i doubled everything

- anonymous

ok so now we have
\[5-2\epsilon

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

8 1/2

- anonymous

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

- anonymous

ok truthfully i dont understand what u did

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

ok

- anonymous

thank you!

- anonymous

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
\[|2.x+6-8.5|<\epsilon\]

- anonymous

typo

- anonymous

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

- anonymous

i will write it if you like

- anonymous

i dont undestand what your asking me to do

- anonymous

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

- anonymous

okat

- anonymous

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

- anonymous

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

- anonymous

hope this helps

- anonymous

wow THANK YOU SO MUCH

- anonymous

welcome.

- anonymous

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

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