How do you use the epsilon delta definition to prove that the limit of #x^2-7x+3=-7# as #x->2#?

The preliminary analysis is a bit long. If you just want to read the proof, scroll down.

Preliminary analysis

By definition,

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

Now we need to actually write up the proof:

Proof

To use the epsilon-delta definition to prove that the limit of the function f(x) = x^2 - 7x + 3 is -7 as x approaches 2, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x - 2| < delta, then |f(x) - (-7)| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a suitable delta > 0. We start by considering the expression |f(x) - (-7)| and try to manipulate it to obtain an expression involving |x - 2|.

|f(x) - (-7)| = |x^2 - 7x + 3 + 7| = |x^2 - 7x + 10|

Now, we want to find a delta such that if 0 < |x - 2| < delta, then |x^2 - 7x + 10| < epsilon.

To simplify the expression further, we can factorize the quadratic:

x^2 - 7x + 10 = (x - 2)(x - 5)

Now, we can rewrite the expression as:

|x - 2||x - 5|

Since we are interested in the behavior of x as it approaches 2, we can assume that |x - 2| < 1 (or any other suitable value). This allows us to establish an upper bound for |x - 5|.

|x - 2| < 1 implies -1 < x - 2 < 1, which leads to 1 < x < 3.

From this, we can deduce that |x - 5| < 2.

Now, we can rewrite the expression as:

|x - 2||x - 5| < 2|x - 2|

We want to ensure that 2|x - 2| < epsilon. Therefore, we can choose delta = min(1, epsilon/2).

If we assume 0 < |x - 2| < delta, then it follows that |x - 2| < 1 and 2|x - 2| < 2.

Since delta = min(1, epsilon/2), we have 2|x - 2| < epsilon.

Thus, we have shown that for any given epsilon > 0, there exists a delta > 0 (specifically, delta = min(1, epsilon/2)) such that if 0 < |x - 2| < delta, then |f(x) - (-7)| < epsilon.

Therefore, by the epsilon-delta definition, the limit of f(x) = x^2 - 7x + 3 as x approaches 2 is -7.