- #1
Hollysmoke
- 185
- 0
I'm supposed to prove that lim x -> -2, x^2+3x+7 = 5
Here's what I have:
|x^2 +3x+7 – 5| < ε
|x+2| < δ
|x^2+3x+2| -> |(x+2)(x+1)| < ε whenever |x+2| < δ
|x+1||x+2| < δ |x+1|
|x+1||x+2| < δ|x+1| < ε
ε / |x+1| > δ , as x -> -2, |x+1| -> 3,
therefore: ε / -1 > δ
But I'm not sure if I did it correctly, since its our first time using quadratic. Help, please?
Here's what I have:
|x^2 +3x+7 – 5| < ε
|x+2| < δ
|x^2+3x+2| -> |(x+2)(x+1)| < ε whenever |x+2| < δ
|x+1||x+2| < δ |x+1|
|x+1||x+2| < δ|x+1| < ε
ε / |x+1| > δ , as x -> -2, |x+1| -> 3,
therefore: ε / -1 > δ
But I'm not sure if I did it correctly, since its our first time using quadratic. Help, please?