# How do you use the epsilon delta definition to prove that the limit of #2x-4=6# as #x->1#?

It cannot be proven. It is false.

We have shown that:

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To use the epsilon-delta definition to prove that the limit of 2x-4 is 6 as x approaches 1, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 1| < delta, then |(2x - 4) - 6| < epsilon.

Let's proceed with the proof:

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

| (2x - 4) - 6 | = |2x - 10| = 2|x - 5|

To ensure that |(2x - 4) - 6| < epsilon, we can set 2|x - 5| < epsilon.

Dividing both sides by 2, we have |x - 5| < epsilon/2.

Now, we can choose delta = epsilon/2.

If 0 < |x - 1| < delta, then |x - 5| < epsilon/2.

Multiplying both sides by 2, we get 2|x - 5| < epsilon.

Since 2|x - 5| = |(2x - 4) - 6|, we have shown that for any epsilon > 0, there exists a delta > 0 such that if 0 < |x - 1| < delta, then |(2x - 4) - 6| < epsilon.

Therefore, by the epsilon-delta definition, the limit of 2x-4 is 6 as x approaches 1.

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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 # (tan2x)/(5x)# as x approaches 0?
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- How do you find the limit of #(sqrt(x+6)-x)/(x^3-3x^2)# as #x->-oo#?
- What is the limit of #(3x^2+20x)/(4x^2+9)# as x goes to infinity?

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