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#.