anonymous
  • anonymous
By the method of delta-epsilon prove limx->5 ((1/2)x+6)= 8 and a1/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
ugh.... epsilons and deltas.....
amistre64
  • amistre64
f(x)-L < e |x-c|
amistre64
  • amistre64
at least its a line and not a curve

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anonymous
  • anonymous
i know all of that sorry let me tell you where am having a problem
anonymous
  • anonymous
i know all of that sorry let me tell you where am having a problem
anonymous
  • anonymous
ok i simplified it to get 1/2 (x-2) < e so how do i find my delta from there?
amistre64
  • amistre64
|x-5| < d right? id have to work it out meself to recall how to do it lol
anonymous
  • anonymous
yes
anonymous
  • anonymous
yes
anonymous
  • anonymous
its |x-5|
amistre64
  • amistre64
a1/2 ?? whats that mean? -e < (1/2)x+6) - 8 < e
anonymous
  • anonymous
8 and a half
amistre64
  • amistre64
-e < (1/2)x-2 < e 2 -e < (1/2)x < 2+e 2(2 -e) < x < 2(2+e)
amistre64
  • amistre64
0 < x-5
amistre64
  • amistre64
5+d < 2(2+e) i got know idea if this is working lol
anonymous
  • anonymous
arent we suppose to be converting x-2 to x-5?..no?
anonymous
  • anonymous
usually you work these problems backwards.
anonymous
  • anonymous
this one is tricky...
anonymous
  • anonymous
start with | \[|.5x+6-8.5|<\epsilon\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[-\epsilon < .5x-2.5< \epsilon\]
amistre64
  • amistre64
ohh....8' 1/2 ....
anonymous
  • anonymous
\[2.5-\epsilon<.5x<2.5+\epsilon\]
anonymous
  • anonymous
\[5-2\epsilon
amistre64
  • amistre64
looks familiar so far lol
anonymous
  • anonymous
can u do it in terms of epsilon and delta?
anonymous
  • anonymous
can u do it in terms of epsilon and delta?
anonymous
  • anonymous
how am i doing so far?
anonymous
  • anonymous
good
amistre64
  • amistre64
well, other than having the right numbers; looks the same as mine lol
anonymous
  • anonymous
oh my delta will depend on epsilon. i am just trying to figure out what the delta needs tgo be
anonymous
  • anonymous
kk
anonymous
  • 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
  • anonymous
ok but in class we normally would deleta write where u wrote your epsilon but its problem a different example
anonymous
  • anonymous
wait
anonymous
  • anonymous
how'd you get the .5 to 5?
anonymous
  • 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
  • anonymous
the 5?
anonymous
  • anonymous
we had \[2.5- \epsilon<.5x<2.5+\epsilon\]
anonymous
  • anonymous
to get x by itself i doubled everything
anonymous
  • anonymous
ok so now we have \[5-2\epsilon
anonymous
  • anonymous
this says the same thing as \[|x-5|<2\epsilon\]
anonymous
  • anonymous
so the delta i pick is \[2\epsilon\]
anonymous
  • anonymous
oh ok sorry i had to move from the pc a sec
anonymous
  • 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
  • anonymous
don't forget what the goal is. you want to show that \[lim_{x->5} (.5x+6)=8\]
anonymous
  • anonymous
8 1/2
anonymous
  • 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
  • anonymous
ok truthfully i dont understand what u did
anonymous
  • anonymous
i say pick \[\delta = 2 \epsilon\] i win by running the algebra backwards
anonymous
  • 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
  • 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
  • anonymous
i got \[\delta = 2\epsilon\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
thank you!
anonymous
  • 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
  • anonymous
typo
anonymous
  • anonymous
i meant start with \[|x-5|<2\epsilon\]
anonymous
  • anonymous
i will write it if you like
anonymous
  • anonymous
i dont undestand what your asking me to do
anonymous
  • anonymous
let me write it and maybe it will make sense. hold on and i will write it all out.
anonymous
  • anonymous
okat
anonymous
  • 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
  • anonymous
sorry it took so long to write. but i never would have found delta = 2 epsilon without working the problem backwards
anonymous
  • anonymous
hope this helps
anonymous
  • anonymous
wow THANK YOU SO MUCH
anonymous
  • anonymous
welcome.
anonymous
  • 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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