anonymous 5 years ago Prove the statement using the epsilon, sigma definition of limit. lim (7-3x)=-5 x--->4

1. anonymous

I am struggling with these. :( Would be nice if some1 could show it.

2. anonymous

i have no idea

3. 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

4. anonymous

but what does that equation mean and how do you solve it

5. anonymous

good question :D I am looking at my notes now and trying to figure it out but not working well

6. anonymous

me too

7. anonymous

I am doing real analysis, are u too?

8. anonymous

no

9. anonymous

I'll be back in 10 min and give it another go

10. anonymous

So the definition is $given any \epsilon >0 there \exists \delta >0 such that 0<\left| x-a \right|<\delta \implies \left| \left| f(x)-L \right| \right|<\epsilon$

11. anonymous

a is 4 L is -5 here

12. anonymous

f(x)=7-3x

13. 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

14. anonymous

Yes that is fine, but it does not use epsilon or delta...

15. anonymous

I guess here you need to use this definition but I'm not sure how. :(

16. anonymous

i give up

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