Consider the line f(x)=mx+b, where m does not equal 0, how do you use the epsilon delta definition of a limit to prove that the limit f(x)=mc+b as x approaches c?

1 Answer
Jim H
Dec 5, 2016

Choose #delta <= epsilon/abs(m)#.

Explanation:

The if #0 < abs(x-c) < delta#, then we have

#abs(f(x) - (mc+b)) = abs((mx+b)-(mc+b))#

# = abs(mx-mc)#

# = abs(m)abs(x-c)#

# < abs(m)delta# #" "# (see Note below)

# <= abs(m) epsilon/abs(m)#

# = epsilon#

That is:

if #0 < abs(x-c) < delta#, then #abs(f(x) - (mc+c)) < epsilon#.

Therefore, by the definition of limit,

#lim_(xrarrc)(mx+b) = mc+b#

Note : Given #abs(x-c) < delta#, we can multiply on both sides by #absm# -- which is positive -- to get #abs(m)abs(x-c) < abs(m)delta#

Also Note that we must use the most strict inequality between the beginning and the end to connect the first and last expressions.

Final also note: If #m = 0#, then the result is still true, but the proof does not work, because we cannot use #epsilon/absm#. (If #m = 0#, use #delta = # any number you like.)

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