# How do you prove that the #lim_(x to -2)(x^2 -5) = -1# using the formal definition of a limit?

We want to show:

Please see the details below.

and

#Rightarrow|x+2|<1 Rightarrow-1 < x+2 < 1 Rightarrow-5< x-2 <-3 Rightarrow |x-2|<5#

So, we have

I hope that this was clear.

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To prove that the limit of (x^2 - 5) as x approaches -2 is equal to -1 using the formal definition of a limit, we need to show that for any given ε > 0, there exists a δ > 0 such that if 0 < |x - (-2)| < δ, then |(x^2 - 5) - (-1)| < ε.

Let's proceed with the proof:

Given ε > 0, we need to find a δ > 0 such that if 0 < |x - (-2)| < δ, then |(x^2 - 5) - (-1)| < ε.

First, let's simplify the expression |(x^2 - 5) - (-1)|:

|(x^2 - 5) - (-1)| = |x^2 - 5 + 1| = |x^2 - 4| = |(x - 2)(x + 2)|

Now, we want to find a δ > 0 such that if 0 < |x - (-2)| < δ, then |(x - 2)(x + 2)| < ε.

Let's consider the term |x + 2|:

|x + 2| = |(x - (-2)) + 4| = |x + 6|

We can see that if we choose δ = 1, then for any x satisfying 0 < |x - (-2)| < 1, we have:

|x + 6| < |x - (-2)| + 6 < 1 + 6 = 7

Now, let's choose δ = min{1, ε/7}. If 0 < |x - (-2)| < δ, then:

|x + 6| < 7

Since |(x - 2)(x + 2)| = |x + 6|, we have:

|(x - 2)(x + 2)| < 7

Since 7 > ε, we have shown that for any given ε > 0, there exists a δ > 0 such that if 0 < |x - (-2)| < δ, then |(x^2 - 5) - (-1)| < ε.

Therefore, we have proven that the limit of (x^2 - 5) as x approaches -2 is equal to -1 using the formal definition of a limit.

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