Use the definition of limits to show that
Lim x➡ 0 : (x^2 - 9) / (x + 3) = -6
hence find the number d > 0 given that e=0.05
for all x ≠ -3, (x^2-9)/(x+3) = x-3
So, the limit as x->0 is just -3.
How did you get -6?
To show that Lim x➡ 0 : (x^2 - 9) / (x + 3) = -6 using the definition of limits, we need to demonstrate that for any given value ε > 0, there exists a value δ > 0 such that |(x^2 - 9) / (x + 3) - (-6)| < ε whenever 0 < |x| < δ.
Let's begin by working on the left-hand side of the inequality:
|(x^2 - 9) / (x + 3) - (-6)|
First, simplify the expression inside the absolute value:
|(x^2 - 9 + 6(x + 3)) / (x + 3)|
Next, combine like terms:
|(x^2 + 6x + 9) / (x + 3)|
Now, let's analyze this expression further. Notice that when x = -3, the denominator becomes zero, resulting in an undefined value. Therefore, we need to evaluate the expression for x ≠ -3.
To simplify the expression, we can factor the numerator:
|(x + 3)(x + 3) / (x + 3)|
Now, divide out the common factor (x + 3):
|x + 3|
To satisfy the inequality |(x + 3)| < δ, we can set δ = ε.
Therefore, for any ε > 0, if 0 < |x| < δ (in this case, if 0 < |x + 3| < ε), we have:
|(x^2 - 9) / (x + 3) - (-6)| < ε
This shows that Lim x➡ 0 : (x^2 - 9) / (x + 3) = -6.
Now, let's move on to finding the value of d > 0 given e = 0.05.
We want to find a value of δ such that whenever |x| < δ (or equivalently, |x + 3| < δ), the expression |(x^2 - 9) / (x + 3) - (-6)| is less than 0.05.
Using the previous analysis, we know that |(x^2 - 9) / (x + 3) - (-6)| = |x + 3|. So we need to find δ such that |x + 3| < 0.05 whenever |x + 3| < δ.
In this case, we can set δ = 0.05.
So, for any x, if |x + 3| < 0.05, then |(x^2 - 9) / (x + 3) - (-6)| = |x + 3| < 0.05. We have found the value of δ.