How do you prove the statement lim as x approaches 2 for # (x^2 - 3x) = -2# using the epsilon and delta definition?

Answer 1

See the explanation, below.

Initial Analysis

We want to make #abs((x^2-3x)-(-2)) < epsilon#
By choosing #delta#, we control the maximum size of #abs(x-2)#
We now look at #abs((x^2-3x)-(-2))# keeping in mind what we control.
#abs((x^2-3x)-(-2)) = abs(x^2-3x+2)#
# = abs((x-1)(x-2))#
# = abs(x-1)abs(x-2)#
If we knew the size of #abs(x-1)#, we could choose #abs(x-2)# to make sure that the product is #< epsilon#
Let's start by making sure that #x# is a little close to #2#, say #abs(x-2) < 1# (#delta# will need to be at most #1#.)
If #abs(x-2) < 1#, then #-1 < x-2 < 1#, that is: #1 < x < 3#.
If follows that #0 < x-1 < 2#, so we have #abs(x-1) < 2#
Now if we ALSO make sure that #abs(x-2) < epsilon/2#,
then we will have #abs(x-1)abs(x-2) < 2abs(x-2) < 2 (epsilon/2) = epsilon#

We are now prepared to write the evidence:

Proof

Given #epsilon > 0#, let #delta = min{1,epsilon/2}#. (Note that #delta is positive.)
For every #x# that satisfies #0 < abs(x-2) < delta# we have
(#abs(x-1) < 2# and also)
#abs((x^2-3x)-(-2)) = abs(x^2-3x+2)#
# = abs((x-1)(x-2))#
# = abs(x-1)abs(x-2)#
# < (2)(epsilon/2) = epsilon#
We have shown that for any #epsilon > 0# there is a #delta > 0# such that
if #0 < abs(x-2) < delta#, then #abs((x^2-3x)-(-2)) < epsilon#.
By the definition of limit, #lim_(xrarr2)(x^2-3x) = -2#.
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Answer 2

To prove the statement lim as x approaches 2 for (x^2 - 3x) = -2 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 0 < |x - 2| < delta, then |(x^2 - 3x) - (-2)| < epsilon.

Let's proceed with the proof:

Given the function f(x) = x^2 - 3x, we want to find a delta such that if 0 < |x - 2| < delta, then |(x^2 - 3x) - (-2)| < epsilon.

First, let's simplify the expression |(x^2 - 3x) - (-2)|: |(x^2 - 3x) + 2| = |x^2 - 3x + 2|

Now, we can factorize the expression inside the absolute value: |x^2 - 3x + 2| = |(x - 2)(x - 1)|

Since we are interested in the behavior of the function as x approaches 2, we can assume that delta is small enough such that 0 < |x - 2| < delta implies 0 < |x - 2| < 1. This allows us to make the following observations:

  1. If 0 < x < 3, then 0 < x - 1 < 2.
  2. If 1 < x < 3, then 0 < x - 2 < 1.

Using these observations, we can rewrite the expression |(x - 2)(x - 1)| as follows:

  1. If 0 < x < 3, then |(x - 2)(x - 1)| = |x - 2||x - 1|.
  2. If 1 < x < 3, then |(x - 2)(x - 1)| = |x - 2||x - 1| = -(x - 2)(x - 1).

Now, let's consider the case when 0 < x < 3:

|x - 2||x - 1| = |x - 2||x - 1| < 2|x - 2|

We can choose delta = min(1, epsilon/2). Then, if 0 < |x - 2| < delta, we have:

|x - 2||x - 1| < 2|x - 2| < 2(delta) = 2(min(1, epsilon/2)) = epsilon

Therefore, we have shown that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x - 2| < delta, then |(x^2 - 3x) - (-2)| < epsilon. This proves the statement lim as x approaches 2 for (x^2 - 3x) = -2 using the epsilon and delta definition.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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