How do you prove the statement lim as x approaches 2 for #(x^2 - 4x + 5) = 1# using the epsilon and delta definition?

Question

Emma Aldridge · Accepted Answer

Please see below.

The explanation is divided into two sections: first, a preliminary analysis is presented to determine the values utilized in the proof, and then the proof itself is presented. locating the evidence 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#. We are required to demonstrate that #lim_(xrarrcolor(green)(2))color(red)(x^2-4x+5) = color(blue)(1)# So we want to make #abs(underbrace(color(red)(x^2-4x+5))_(color(red)(f(x)) )-underbrace(color(blue)(1))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((2)))_color(green)(a))# We want: #abs((x^2-4x+5)-1) < epsilon# Examine the object we wish to reduce in size. Edit this, seeking the object under our control. #abs((x^2-4x+5)-1) =abs(x^2-4x+4)# # = abs((x-2)^2) # # = (x-2)^2# In order to make this less than #epsilon#, it suffices to make #abs(x-2)# less than #sqrtepsi# Writing the proof to demonstrate that our L is accurate Claim: #lim_(xrarr2)(x^2-4x+5) = 1# Proof: Given #epsilon > 0#, choose #delta = sqrtepsilon#. (Note that #delta# is positive.) Now if #0 < abs(x-2) < delta# then #abs((x^2-4x+5)-1) =abs(x^2-4x+4)# # = abs((x-2)^2) # # = (x-2)^2# # < delta^2# #" "# (See Note below) # = (sqrtepsilon)^2# # = epsilon# We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-2) < delta#, then #abs((x^2-4x+5)-1) < epsilon#. So, by the definition of limit, we have #lim_(xrarr2)(x^2-4x+5) = 1#. Note Since positive values cause the squaring function to increase, #abs(x-2) < sqrtdelta# implies that #abs(x-2)^2 = (x-2)^2 < delta^2#

Ezekiel Alexander · Answer

undefined To prove the statement using the epsilon and delta definition, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if the distance between x and 2 is less than delta, then the distance between (x^2 - 4x + 5) and 1 is less than epsilon.

Let's start by finding the expression for (x^2 - 4x + 5):

(x^2 - 4x + 5) = (x - 2)^2 + 1

Now, let's consider the distance between (x^2 - 4x + 5) and 1:

|(x^2 - 4x + 5) - 1| = |(x - 2)^2 + 1 - 1| = |(x - 2)^2|

We want to make this distance less than epsilon. So, we can set a condition:

|(x - 2)^2| < epsilon

Now, let's consider the distance between x and 2:

|x - 2|

We want to make this distance less than delta. So, we can set another condition:

|x - 2| < delta

To prove the statement, we need to find a suitable delta in terms of epsilon. We can start by assuming that delta is less than 1, which implies:

|x - 2| < 1

Now, let's manipulate the expression for |(x - 2)^2|:

|(x - 2)^2| = |x - 2|^2 = (|x - 2|)(|x - 2|)

Since |x - 2| < 1, we can say:

|(x - 2)^2| < (1)(|x - 2|) = |x - 2|

We want this to be less than epsilon, so we can set:

|x - 2| < epsilon

Therefore, if we choose delta to be the minimum of 1 and epsilon, we can satisfy both conditions:

If |x - 2| < delta, then |(x - 2)^2| < epsilon

Hence, we have proven the statement using the epsilon and delta definition.