# How do you prove that the limit of #f(x) =2x-3# as x approaches 5 is 7 using the epsilon delta proof?

To prove it using the limit definition consider the value:

which proves the limit.

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Please see below.

# = 2(epsilon/2)

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To prove that the limit of f(x) = 2x - 3 as x approaches 5 is 7 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 0 < |x - 5| < delta, then |f(x) - 7| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 5| < delta, then |f(x) - 7| < epsilon.

We start by manipulating the expression |f(x) - 7| < epsilon:

|f(x) - 7| = |(2x - 3) - 7| = |2x - 10| = 2|x - 5|

Now, we can rewrite the inequality as 2|x - 5| < epsilon.

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

Therefore, we can choose delta = epsilon/2.

Now, if 0 < |x - 5| < delta, then |x - 5| < epsilon/2.

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

Since 2|x - 5| = |f(x) - 7|, we have shown that if 0 < |x - 5| < delta, then |f(x) - 7| < epsilon.

Hence, we have proven that the limit of f(x) = 2x - 3 as x approaches 5 is 7 using the epsilon-delta proof.

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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.

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