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#.
