Am I doing the steps correctly in this epsilon-delta proof?

Question

epsilon finding delta $$\lim_{x \to -1}(3-4x)=7$$ $$|3-4x-7|<ε$$ $$-ε<3-4x-7<ε$$ $$-ε<-4x-4<ε$$ $$4-ε<-4x<ε+4$$ $$-\fracε4 <-x<\fracε4$$ This point here and down, I feel like I am missing a step to remove the negative from the x? Which I think would flip the logic to $\fracε4>x>-\fracε4$ $$|x|<\fracε4=δ$$

delta proving epsilon

$$|x-1|<\fracε4$$ $$-\fracε4<x-1<\fracε4$$ $$1-\fracε4<x<\fracε4+1$$ $$4-ε<-4x<ε+4$$ $$-ε<-4x-4<ε$$ $$-ε<3-4x-7<ε$$ $$|3-4x-7|<ε$$

Answer 1

OK, so at looking it again (and with all the helpful tips), and the use of the distributive property..

epsilon to delta

$$\lim_{x \to -1}(3-4x)=7$$ $$|3-4x-7|<ε$$ $$-ε<3-4x-7<ε$$ $$-ε<-4x-4<ε$$ $$-ε<4(-x-1)<ε$$ $$-\fracε4<-x-1<\fracε4$$ $$1-\fracε4 <-x<\fracε4+1$$ $$\fracε4>x-(-1)>\fracε4$$ $$|x-(-1)|<\fracε4$$ $$|x+1|<\fracε4$$

Answer 2

Where you left:

$$-\frac\epsilon4<-x<\frac\epsilon4\implies -\frac\epsilon4<x<\frac\epsilon4\iff|x|<\frac\epsilon4$$

so choose $\;\delta=\frac\epsilon4\;$ and reverse the arrows in your work (first part) ...why can you? If you want around $\;-1\;$ then

$$-\frac\epsilon4<x<\frac\epsilon4\implies1-\frac\epsilon4<x+1<1+\frac\epsilon4$$

so that you can focus on a neighborhood of $\;-1\;$ (fill in gaps)