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anonymous
 5 years ago
By the method of deltaepsilon prove
limx>5 ((1/2)x+6)= 8 and a1/2
anonymous
 5 years ago
By the method of deltaepsilon prove limx>5 ((1/2)x+6)= 8 and a1/2

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ugh.... epsilons and deltas.....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at least its a line and not a curve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know all of that sorry let me tell you where am having a problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know all of that sorry let me tell you where am having a problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i simplified it to get 1/2 (x2) < e so how do i find my delta from there?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x5 < d right? id have to work it out meself to recall how to do it lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a1/2 ?? whats that mean? e < (1/2)x+6)  8 < e

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0e < (1/2)x2 < e 2 e < (1/2)x < 2+e 2(2 e) < x < 2(2+e)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00 < x5 <d 5 < x < 5+d

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05+d < 2(2+e) i got know idea if this is working lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0arent we suppose to be converting x2 to x5?..no?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0usually you work these problems backwards.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this one is tricky...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0start with  \[.5x+68.5<\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\epsilon < .5x2.5< \epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2.5\epsilon<.5x<2.5+\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[52\epsilon<x<5+2\epsilon\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0looks familiar so far lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can u do it in terms of epsilon and delta?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can u do it in terms of epsilon and delta?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how am i doing so far?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, other than having the right numbers; looks the same as mine lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh my delta will depend on epsilon. i am just trying to figure out what the delta needs tgo be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok but in class we normally would deleta write where u wrote your epsilon but its problem a different example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how'd you get the .5 to 5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am working the problem backwards. then the proof is to work it forwards. you say for example pick delta such that \[52\epsilon < \delta< 5+2\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we had \[2.5 \epsilon<.5x<2.5+\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to get x by itself i doubled everything

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so now we have \[52\epsilon <x < 5+2\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this says the same thing as \[x5<2\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the delta i pick is \[2\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok sorry i had to move from the pc a sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because that is what we wanted. we wanted to say that we could find a delta such that if \[x5<\delta\] then \[.5x6+8.5<\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't forget what the goal is. you want to show that \[lim_{x>5} (.5x+6)=8\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0meaning that given any epsilon > 0 y ou can find a delta such that if \[x5<\delta\] then \[.5x+68.5<\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok truthfully i dont understand what u did

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i say pick \[\delta = 2 \epsilon\] i win by running the algebra backwards

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0delta will be an expression in epsilon. so you start with the thing that you want, namely \[.5x+68.5<\epsilon\] do some algebra, and see what delta you get.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got \[\delta = 2\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0run the algebra forward and see that it works. just take the steps i wrote and write them backwards. in other words start with \[x5<2\delta\] and see that you get \[2.x+68.5<\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i meant start with \[x5<2\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i will write it if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont undestand what your asking me to do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me write it and maybe it will make sense. hold on and i will write it all out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You want to show that given any \[\epsilon>0\] you can find a \[\delta\] such that if \[x5<\delta\] then \[.5x+68.5<\epsilon\] let \[\delta=2\epsilon\] then \[x5<2\epsilon\] means \[2\epsilon<x5<\epsilon\] giving \[52\epsilon < x < 5+2\epsilon\] gives \[2.5\epsilon < .5x < 2.5 +\epsilon\] so \[\epsilon < .5x2.5<\epsilon\] or \[.5x2.5<\epsilon\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry it took so long to write. but i never would have found delta = 2 epsilon without working the problem backwards

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow THANK YOU SO MUCH

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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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