How do you use the epsilon delta definition to prove that the limit of #x/(6-x)=1# as #x->3#?

This would not be necessary as:

However, as an exercise, we can see that:

We have:

and

Putting this two inequalities together:

that is:

we will have:

To prove that the limit of x/(6-x) is 1 as x approaches 3 using the epsilon-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 - 3| < delta, then |x/(6-x) - 1| < epsilon.

Let's begin the proof:

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

|x/(6-x) - 1| = |x - 6 + x| / |6 - x| = |2x - 6| / |6 - x|

To simplify further, we can assume that delta < 1, which implies 2 < x < 4.

Now, let's consider |2x - 6| / |6 - x| < epsilon.

Since 2 < x < 4, we can say that 0 < 6 - x < 2.

Therefore, |2x - 6| / |6 - x| < |2x - 6| / 2.

To ensure that |2x - 6| / 2 < epsilon, we can set |2x - 6| < 2epsilon.

Now, we can choose delta = min{1, 2epsilon}.

If 0 < |x - 3| < delta, then it implies that 2 < x < 4 and |2x - 6| < 2epsilon.

Hence, |x/(6-x) - 1| = |2x - 6| / |6 - x| < |2x - 6| / 2 < 2epsilon / 2 = epsilon.

Therefore, we have shown that for any epsilon > 0, there exists a delta > 0 such that if 0 < |x - 3| < delta, then |x/(6-x) - 1| < epsilon.

Hence, the limit of x/(6-x) as x approaches 3 is indeed 1.