How do you prove that the limit of # (x²-9) / (x²+5x+6)=0 # as x approaches -3 using the epsilon delta proof?

First we note that:

So the function can be simplified as:

To prove that the limit of (x²-9) / (x²+5x+6) is 0 as x approaches -3 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that whenever 0 < |x - (-3)| < delta, then |(x²-9) / (x²+5x+6) - 0| < epsilon.

Let's begin by simplifying the expression (x²-9) / (x²+5x+6). Factoring the numerator and denominator, we get (x-3)(x+3) / (x+2)(x+3). Notice that (x+3) cancels out, leaving us with (x-3) / (x+2).

Now, let's proceed with the epsilon-delta proof. We want to find a delta such that whenever 0 < |x - (-3)| < delta, then |(x-3) / (x+2) - 0| < epsilon.

First, let's consider the denominator (x+2). Since we want to approach -3, we need to ensure that x+2 is not equal to 0. Therefore, we can set a condition that delta is less than 2, ensuring that x+2 will never be 0.

Next, let's focus on the numerator (x-3). We want to find a delta such that whenever 0 < |x - (-3)| < delta, then |(x-3) / (x+2) - 0| < epsilon. We can start by assuming that delta is less than 1, which implies -4 < x < -2.

Now, let's manipulate the expression |(x-3) / (x+2) - 0| to simplify it further. We have |(x-3) / (x+2)|. Since we know -4 < x < -2, we can conclude that -1 < (x-3) / (x+2) < 1.

To ensure that |(x-3) / (x+2) - 0| < epsilon, we can set a condition that delta is less than min{1, epsilon}. This guarantees that whenever 0 < |x - (-3)| < delta, then |(x-3) / (x+2) - 0| < epsilon.

Therefore, we have proven that the limit of (x²-9) / (x²+5x+6) is 0 as x approaches -3 using the epsilon-delta proof.