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.
