How do you prove the statement lim as x approaches 1 for #(5x^2)=5# using the epsilon and delta definition?

We need to make #abs(5x^2-5) < epsilon# by making #abs(x-1) < delta#. (We choose the #delta#.)

So we begin by examining #abs(5x^2-5)#

#abs(5x^2-5) = abs(5(x^2-1)) = abs5abs(x^2-1)#

# = 5 abs((x+1)(x-1)) = 5 abs(x+1) abs(x-1)#

We control, through our choice of #delta#, the maximum for #abs(x-1)#.

If we make sure that #delta# is less than a number we choose now, then we can also control the size of #abs(x+1)#. (If we put a bound on the distance between #x# and #1#, the we also get a bound on the distance between #x# and #-1#)

Let's make sure that #delta <= 2#.

If #abs(x-1) < 2#, then #-2 < x-1 < 2#.

So, adding 2 to each part, we get, #0 < x+1 < 4# and #abs(x+1) < 4#.

We want #5 abs(x+1) abs(x-1) < epsilon# and we plan to make sure that #abs(x+1) < 4#, so we now need

#5 abs(x+1) abs(x-1) #

#"which is" < 5(4)abs(x-1)# # "which is"= 20 abs(x-1)# #"we want this" < epsilon#

So let's make sure that, in addition to #delta <= 2#,

we also want #delta <= epsilon/20#

Proof Claim: #lim_(xrarr1)5x^2 = 5#

Given #epsilon > 0#, let #delta = min{2, epsilon/20}# (Note that #delta > 0# as required.)

If #x# is chosen so that #0 < abs(x-1) < delta# then

note first that #abs(x+1) < 4# ( #abs(x-1) < delta <= 2 rArr -2 < x-1 < 2 rArr 0 < x+1 < 4#).

Furthermore, for such #x#, we have

#abs(5x^2-5) = 5abs(x^2-1) = 5abs(x+1)abs(x-1)#

# < 5(4) abs(x-1) = 20abs(x-1)#

# < 20(epsilon/20) = epsilon#.

That is: for #0 < abs(x-1< delta#, we have #abs(5x^2-5) < epsilon.

# lim_(xrarr1)5x^2 = 5#

To prove the statement lim as x approaches 1 for (5x^2) = 5 using the epsilon and delta definition, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that if 0 < |x - 1| < delta, then |(5x^2) - 5| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 1| < delta, then |(5x^2) - 5| < epsilon. First, let's manipulate the expression |(5x^2) - 5|: |(5x^2) - 5| = |5(x^2 - 1)| = 5|x - 1||x + 1| Now, we can see that if we choose delta = min(1, epsilon/10), then the following holds: 0 < |x - 1| < delta implies |x - 1| < 1 and |x - 1| < epsilon/10. Since |x - 1| < 1, we can also conclude that -1 < x - 1 < 1, which implies 0 < x < 2. Now, let's consider the expression |(5x^2) - 5|: 5|x - 1||x + 1| < 5(epsilon/10)(2 + 1) = epsilon/2 Therefore, if 0 < |x - 1| < delta, then |(5x^2) - 5| < epsilon/2. Now, let's consider the expression |(5x^2) - 5| again: |(5x^2) - 5| = 5|x - 1||x + 1| Since 0 < x < 2, we can conclude that 0 < x + 1 < 3. Therefore, 5|x - 1||x + 1| < 5(epsilon/10)(3) = 3epsilon/2. So, if 0 < |x - 1| < delta, then |(5x^2) - 5| < 3epsilon/2. Now, we need to choose a delta that satisfies both conditions: We can choose delta = min(1, epsilon/10). Therefore, if 0 < |x - 1| < delta, then |(5x^2) - 5| < epsilon/2 and |(5x^2) - 5| < 3epsilon/2. Combining these inequalities, we have: |(5x^2) - 5| < epsilon/2 and |(5x^2) - 5| < 3epsilon/2. Since epsilon/2 < 3epsilon/2, we can conclude that |(5x^2) - 5| < epsilon for any epsilon > 0, if 0 < |x - 1| < delta. Therefore, we have proven that lim as x approaches 1 for (5x^2) = 5 using the epsilon and delta definition.