How do you prove that the limit of #(x^2 - x) = 0 # as x approaches 1 using the epsilon delta proof?

Question

Isaiah Allen · Accepted Answer

See the explanation section below.

Preliminary analysis We want to show that #lim_(xrarr1)(x^2-x) = 0#. By definition, #lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)# if and only if for every #epsilon > 0#, there is a #delta > 0# such that: for all #x#, #" "# if #0 < abs(x-color(green)(a)) < delta#, then #abs(color(red)(f(x))-color(blue)(L)) < epsilon#. So we want to make #abs(underbrace(color(red)((x^2-x)))_(color(red)(f(x)) )-underbrace(color(blue)(0))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)(1))_color(green)(a))# Look at the thing we want to make small. Factor this, looking for the thing we control. #abs((x^2-x)-0) = abs(x(x-1)) = absx abs(x-1)# And there's #abs(x-1)#, the thing we control We can make #absx abs(x-1) < epsilon# by making #abs(x-1) < epsilon/absx#, BUT we need a #delta# that is independent of #x#. Here's how we can work around that. If we make sure that the #delta# we eventually choose is less than orequal to #1#, then for every #x# with #abs(x-1) < delta#, we will have #abs(x-1) < 1# which is true if and only if #-1 < x-1 < 1 # which is true if and only if #0 < x < 2#. Consequently: if #abs(x-1) < 1#, then #absx < 2# If we also make sure that #delta <= epsilon/2#, then we will have: for all #x# with #abs(x-1) < delta# we have #abs(x^2-x) = absx abs(x-1)<2(epsilon/2) = epsilon# So we will choose #delta = min{1, epsilon/2}#. (Any lesser #delta# would also work.) Now we need to actually write up the proof: Proof Given #epsilon > 0#, choose #delta = min{1, epsilon/2}#. #" "# (note that #delta# is also positive). Now for every #x# with #0 < abs(x-1) < delta#, we have #absx < 2# and #abs(x-1) < epsilon/2#. So, #abs((x^2-x) - 0) = absx abs(x-1)) <= 2abs(x-1) < 2delta <= 2 epsilon/2 = epsilon# Therefore, with this choice of delta, whenever #0 < abs(x-1) < delta#, we have #abs((x^2-x) - 0) < epsilon# So, by the definition of limit, #lim_(xrarr1)(x^2-x) = 0#.

Adam Adkins · Answer

undefined To prove that the limit of (x^2 - x) as x approaches 1 is 0 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that whenever 0 < |x - 1| < delta, then |(x^2 - x) - 0| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a delta > 0 such that whenever 0 < |x - 1| < delta, then |(x^2 - x) - 0| < epsilon. First, let's simplify |(x^2 - x) - 0| to |x^2 - x|. Now, we can factor |x^2 - x| as |x(x - 1)|. Since we are interested in values of x close to 1, we can assume that |x - 1| < 1, which implies -1 < x - 1 < 1. Using the triangle inequality, we have |x(x - 1)| ≤ |x| |x - 1|. Since we want to find a delta such that |(x^2 - x) - 0| < epsilon, we can set |x| |x - 1| < epsilon. Now, we can choose a delta such that whenever 0 < |x - 1| < delta, we have |x| |x - 1| < epsilon. Let's choose delta = min(1, epsilon). If 0 < |x - 1| < delta, then we have -1 < x - 1 < 1, which implies -1 < x < 2. Since we assumed |x - 1| < 1, we can also assume that 0 < x < 2. Now, let's consider |x| |x - 1| < epsilon. Since 0 < x < 2, we have |x| < 2. Therefore, |x| |x - 1| < 2 |x - 1|. Since we chose delta = min(1, epsilon), we have 2 |x - 1| < 2 delta ≤ 2 epsilon. Hence, whenever 0 < |x - 1| < delta, we have |(x^2 - x) - 0| = |x| |x - 1| < 2 epsilon. This completes the epsilon-delta proof, showing that the limit of (x^2 - x) as x approaches 1 is indeed 0.