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?

Question

Paisley Johnson · Accepted Answer

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

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.)

Emily Ammerman · Answer

undefined To use the epsilon-delta definition of a limit to prove that the limit of f(x) as x approaches c is equal to mc + b, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x - c| < delta, then |f(x) - (mc + b)| < epsilon. Let's proceed with the proof: Given the line f(x) = mx + b, we want to show that the limit of f(x) as x approaches c is mc + b. We start by considering |f(x) - (mc + b)| and try to manipulate it to fit the epsilon-delta definition. |f(x) - (mc + b)| = |mx + b - (mc + b)| = |mx - mc| = |m(x - c)| = |m||x - c| Now, we can see that if we choose delta = epsilon/|m|, then for any 0 < |x - c| < delta, we have: |m||x - c| < |m|(epsilon/|m|) = epsilon Thus, we have shown that for any epsilon > 0, there exists a delta > 0 (specifically, delta = epsilon/|m|) such that if 0 < |x - c| < delta, then |f(x) - (mc + b)| < epsilon. Therefore, by the epsilon-delta definition of a limit, we have proven that the limit of f(x) as x approaches c is equal to mc + b.