How do you prove that the limit of #(1-5x) = 16# as x approaches -3 using the epsilon delta proof?

Question

1969 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T02:22:49

See proof below.

Given a positive #epsilon#, let #delta = epsilon/5#.

Now for every #x# with #0 < abs(x-(-3)) < delta#, we have

#0 < abs(x+3) < delta# and

#abs((1-5x)-16) = abs(-5-15)#

# = abs((-5)(x+3))#

# = abs(-5)abs(x+3)#

# = 5abs(x+3)#

# < 5delta = 5(epsilon/5) = epsilon#

We have shown that for any positive #epsilon#, there is a positive #delta# such that for every #x#,

if #0 < abs(x-(-3)) < delta#, then #abs((1-5x)-16) < epsilon#.

Therefore, by the definition of limit, #lim_(xrarr-3)(1-5x) = 16#.