Question

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

Answer 1

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

Answer 2

i have no idea

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

me too

Answer 7

I am doing real analysis, are u too?

Answer 8

no

Answer 9

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

Answer 10

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

Answer 11

a is 4 L is -5 here

Answer 12

f(x)=7-3x

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

i give up