How do you prove the statement lim as x approaches 4 for #(7 – 3x) = -5# using the epsilon and delta definition?

Answer 1

See the explanation section below.

Our initial research

We need to show that, given any positive #edpsilon#, there is a #delta# (also positive) that guarantees that if #x# is chosen so that
#0 < abs(x-4) < delta#, then we will also have #abs((7-3x)-(-5))< epsilon#.
Let's look at the thing we need to make less than #epsilon#.
#abs((7-3x)-(-5)) = abs(7-3x+5)#
# = abs(12-3x)#
Because we say how to find #delta#, that means we control the sive of #abs(x-4)#
Not that if we factor out a #-#, we can get
#abs((7-3x)-(-5)) = abs(12-3x) = abs(-3(x-4))#
# = abs(-3)abs(x-4) = 3abs(x-4)#
We have found that #abs((7-3x)-(-5))# which we want to make less that #epsilon#, is equal to #3abs(x-4)#, which is #3# times the thing we control.
If we make #abx(x-4) < epsilon/4#, that will work.

We now have to provide our work as evidence.

Proof

Given #epsilon > 0#, choose #delta = epsilon/4#. Now if #x# is chosen so that #0 < abs(x-4) < delta#, then we have
#abs((7-3x)-(-5)) = abs(12-3x) = abs(-3)abs(x-4) #
# = 3abs(x-4) < 3 delta = 3(epsilon/3) = epsilon.#.
That is, #abs((7-3x)-(-5)) < epsilon.#.

Take note that we utilized the fact that near the middle, without mentioning it.

#abs(x-4) < delta# implies that #3abs(x-4) < 3delta#.
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Answer 2

To prove the statement lim as x approaches 4 for (7 – 3x) = -5 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 - 4| < delta, then |(7 – 3x) - (-5)| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a suitable delta > 0.

| (7 – 3x) - (-5) | = | 7 - 3x + 5 | = | 12 - 3x |

To ensure that | (7 – 3x) - (-5) | < epsilon, we can set up the following inequality:

| 12 - 3x | < epsilon

Now, we can manipulate the inequality to isolate x:

-epsilon < 12 - 3x < epsilon

Subtracting 12 from all parts of the inequality:

-epsilon - 12 < -3x < epsilon - 12

Dividing all parts of the inequality by -3 (since -3 is negative, the inequality direction changes):

(epsilon + 12) / 3 > x > (12 - epsilon) / 3

Let delta = min((epsilon + 12) / 3, (12 - epsilon) / 3)

Now, if 0 < |x - 4| < delta, we can substitute the value of delta:

0 < |x - 4| < min((epsilon + 12) / 3, (12 - epsilon) / 3)

Since delta is positive, we can multiply all parts of the inequality by 3:

0 < 3|x - 4| < min(epsilon + 12, 12 - epsilon)

Since epsilon is positive, we can simplify the inequality:

0 < 3|x - 4| < epsilon + 12

Now, we can divide all parts of the inequality by 3:

0 < |x - 4| < (epsilon + 12) / 3

Therefore, we have shown that for any given epsilon > 0, there exists a delta > 0 (specifically, delta = min((epsilon + 12) / 3, (12 - epsilon) / 3)) such that if 0 < |x - 4| < delta, then |(7 – 3x) - (-5)| < epsilon. This satisfies the epsilon and delta definition, proving the statement lim as x approaches 4 for (7 – 3x) = -5.

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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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