use &-e to prove the following
lim x^2-9/x+3 = -6
x->-3
please help me i really need help:(
the & stands for delta and e stands for epsilon
To prove the given limit using the epsilon-delta definition, we need to find a value for delta such that the absolute value of x - (-3) is less than delta implies that the absolute value of (x^2 - 9) / (x + 3) - (-6) is less than epsilon.
Let's start by rewriting the expression:
|x^2 - 9| / |x + 3 - (-3)| < ε
First, let's factor the numerator:
|x^2 - 9| = |(x - 3)(x + 3)| = |x - 3| * |x + 3|
Now, we can rewrite the expression:
|x - 3| * |x + 3| / |x + 3 - (-3)| < ε
|x - 3| * |x + 3| / |x + 6| < ε
Since we're looking for a value of delta near x = -3, we can assume that delta is positive. Therefore, we can drop the absolute value signs:
(x - 3) * (x + 3) / (x + 6) < ε
Now, let's work on finding a bound for |x - (-3)| = |x + 3|:
|x + 3| < δ
Next, we can add 6 to both sides:
|x + 9| < δ + 6
Now, we can find a bound for |x + 6|:
|x + 6| ≤ |x + 9| + 3
Substituting this into the previous inequality:
(x - 3) * (x + 3) / (|x + 9| + 3) < ε
Now, we can bound |x - (-3)| = |x + 3| with δ:
|x + 3| < δ
Adding 3 to both sides:
|x + 6| < δ + 3
Using this bound, we can rewrite the expression:
(x - 3) * (x + 3) / (|x + 6|) < ε
Now, we can choose a value for delta such that if |x + 3| < δ, the expression is less than ε. We can do this by manipulating the expression to simplify it further:
|x + 3| = |x - (-3)|
Let's choose a value for delta such that |x - (-3)| < δ:
|x - (-3)| < δ
If we set δ ≤ 1, then |x - (-3)| < δ implies that -1 < (x + 3) < 1, which implies -4 < x < -2.
Therefore, for any ε > 0, if we choose δ ≤ 1 and -4 < x < -2, the expression will be less than ε.
Thus, we have proven the limit using the epsilon-delta definition.