# How do you prove that the limit of #x^2 = 9# as x approaches 3 using the epsilon delta proof?

See below.

The epsilon delta proof for limits is easier understood when one is familiar with the definitions of the terms involved. Most useful will be the definition of the limit of a function.

Definition:

In our case:

So:

We are looking at the maximum values for both cases.

So we know that we must have:

Then our two restrictions are:

Note that all of the above steps come from the definition of the limit of a function as provided above.

Another way to go about working out the scratch work is as follows:

So, for

By signing up, you agree to our Terms of Service and Privacy Policy

To prove that the limit of x^2 is 9 as x approaches 3 using the epsilon-delta proof, we need to show that for any given positive value of epsilon, there exists a positive value of delta such that if the distance between x and 3 (denoted as |x - 3|) is less than delta, then the distance between x^2 and 9 (denoted as |x^2 - 9|) is less than epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 3| < delta, then |x^2 - 9| < epsilon.

We start by assuming 0 < |x - 3| < delta. Expanding the expression |x - 3|, we have -delta < x - 3 < delta.

Adding 3 to all parts of the inequality, we get -delta + 3 < x < delta + 3.

Now, let's consider the expression |x^2 - 9|. Expanding it, we have |x^2 - 9| = |(x - 3)(x + 3)|.

Since we have assumed 0 < |x - 3| < delta, we can also assume that 0 < x - 3 < delta, and thus 0 < x < delta + 3.

Now, we can rewrite |x^2 - 9| as |(x - 3)(x + 3)| < (delta + 3)(x + 3).

To ensure that |x^2 - 9| < epsilon, we need to choose a suitable value for delta. Let's choose delta = min(1, epsilon/6).

Now, if 0 < |x - 3| < delta, we have 0 < x < delta + 3, which implies 0 < x < 1 + 3 = 4.

Using this information, we can rewrite (delta + 3)(x + 3) as (delta + 3)(4 + 3) = 7(delta + 3).

Since we chose delta = min(1, epsilon/6), we have delta ≤ 1, which implies 7(delta + 3) ≤ 7(1 + 3) = 28.

Therefore, if 0 < |x - 3| < delta, we have |x^2 - 9| < 28.

To satisfy the condition |x^2 - 9| < epsilon, we need to ensure that 28 < epsilon.

Hence, by choosing delta = min(1, epsilon/6), we have shown that if 0 < |x - 3| < delta, then |x^2 - 9| < epsilon, proving that the limit of x^2 is 9 as x approaches 3 using the epsilon-delta proof.

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- How do you find the limit of # [ ( t^2 + 2) / (t^3 + t^2 -1) ]# as x approaches negative infinity?
- How do you find the limit of #1/(2x+sinx)# as x approaches #oo#?
- What is the value of #lim_(x->0) (1-cos(mx))/(xsinx)#?
- What is the limit of #((64x^2 +x)^(1/2) -8x)# as x approaches infinity?
- How do you find the limit of #x/sinx# as x approaches 0?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7