anonymous
  • anonymous
Prove the statement using the epsilon, sigma definition of limit. lim (7-3x)=-5 x--->4
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I am struggling with these. :( Would be nice if some1 could show it.
anonymous
  • anonymous
i have no idea
anonymous
  • anonymous
you should start with something like this: for all epsilon (I will use E) >0 there exist a delta (D) such that D>|x-4| implies E>|f(x)-(-5)| Or something like this

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anonymous
  • anonymous
but what does that equation mean and how do you solve it
anonymous
  • anonymous
good question :D I am looking at my notes now and trying to figure it out but not working well
anonymous
  • anonymous
me too
anonymous
  • anonymous
I am doing real analysis, are u too?
anonymous
  • anonymous
no
anonymous
  • anonymous
I'll be back in 10 min and give it another go
anonymous
  • anonymous
So the definition is \[given any \epsilon >0 there \exists \delta >0 such that 0<\left| x-a \right|<\delta \implies \left| \left| f(x)-L \right| \right|<\epsilon\]
anonymous
  • anonymous
a is 4 L is -5 here
anonymous
  • anonymous
f(x)=7-3x
anonymous
  • anonymous
ok so all you do is plug in the 4 for x which gives the answer as 7-12=-5, -5=-5 and that is it. -Source a friend who got an a in calc 3
anonymous
  • anonymous
Yes that is fine, but it does not use epsilon or delta...
anonymous
  • anonymous
I guess here you need to use this definition but I'm not sure how. :(
anonymous
  • anonymous
i give up

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