# Consider the line f(x)=mx+b, where m does not equal 0, how do you use the epsilon delta definition of a limit to prove that the limit f(x)=mc+b as x approaches c?

Choose

That is:

Therefore, by the definition of limit,

Also Note that we must use the most strict inequality between the beginning and the end to connect the first and last expressions.

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To use the epsilon-delta definition of a limit to prove that the limit of f(x) as x approaches c is equal to mc + b, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x - c| < delta, then |f(x) - (mc + b)| < epsilon.

Let's proceed with the proof:

Given the line f(x) = mx + b, we want to show that the limit of f(x) as x approaches c is mc + b.

We start by considering |f(x) - (mc + b)| and try to manipulate it to fit the epsilon-delta definition.

|f(x) - (mc + b)| = |mx + b - (mc + b)| = |mx - mc| = |m(x - c)| = |m||x - c|

Now, we can see that if we choose delta = epsilon/|m|, then for any 0 < |x - c| < delta, we have:

|m||x - c| < |m|(epsilon/|m|) = epsilon

Thus, we have shown that for any epsilon > 0, there exists a delta > 0 (specifically, delta = epsilon/|m|) such that if 0 < |x - c| < delta, then |f(x) - (mc + b)| < epsilon.

Therefore, by the epsilon-delta definition of a limit, we have proven that the limit of f(x) as x approaches c is equal to mc + b.

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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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