How do you prove that the #lim_(xrarr3) (4x-5)=7# using the formal definition of a limit?

Question

7772 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-09T03:11:55

See below.

Given #epsilon > 0#, choose #delta = epsilon/4# (Note that #delta > 0#.)

Then for any #x# with #0 < abs (x-3) < delta#,

#abs((4x-5)-7) = abs(4x-12)#

# = 4abs(x-3)#

# < 4delta#

# = 4epsilon/4#

# = epsilon#

That is: #abs((4x-5)-7) < epsilon#.

We have shown that:

For any #epsilon > 0#, there is a #delta > 0# such that

for all #x# with #0 < abs(x-3) < delta# we have #abs((4x-5)-7) < epsilon#.

So, by the definition of limit, #lim_(xrarr3)(4x-5) = 7#.