How do you find values of δ that correspond to ε=0.1, ε=0.05, and ε=.01 when finding the limit of ##(5x-7)## as x approaches 2?

Make ##delta < epsilon/5##.

We believe that ##lim_(xrarr2)(5x-7)=3##.

To show this, we need to show that we can make ##abs((5x-7)-3)< epsilon## by making ##abs(x-2) < delta##.

We observe that: ##abs((5x-7)-3)=abs(5x-10)=abs(5(x-2))=abs(5)abs(x-2)=5abs(x-2)##.

Makng ##abs((5x-7)-3)< epsilon## can be done by making ##5abs(x-2)< epsilon##.

Which, in turn, can be done by making ##abs(x-2)< epsilon/5##. That is, by making ##delta = epsilon/5##.

For ##epsilon = 0.1##, make ##delta = 0.1/5=0.02##. Now if ##abs(x-2)< delta##, then ##5abs(x-2)## must be < ##5 delta##. (If ##a < b##, then ##5a < 5b##.) So we can be sure that ##abs((5x-7)-3)## which is equal to ##5abs(x-2)## must be < ##5 delta##.

That is ##abs((5x-7)-3)<5(0.02)=0.1##

For ##epsilon = 0.05##, make ##delta = 0.05/5=0.01##. Now if ##abs(x-2)< delta##, then ##5abs(x-2)## must be < ##5 delta##. (If ##a < b##, then ##5a < 5b##.) So we can be sure that ##abs((5x-7)-3)## which is equal to ##5abs(x-2)## must be < ##5 delta##.

That is ##abs((5x-7)-3)<5(0.01)=0.05##

For ##epsilon = 0.01##, make ##delta = 0.01/5=0.002##.