# Thread: Proof question: What's happening?

1. ## Proof question: What's happening?

Okay, so I did get the problem right the first try. However I am not sure what's going on.

I have the problem: Give an epsilon-delta proof of the limit(x^2-5x-3) = 3 as X approaches -1.

At a point in the proof I come to a point where I have ...|(x-6)(x+1)|<E

My notes and book have some really, really wild stuff going on. Basically "Agree that delta is less than or equal to 1", then it does something really wild. The problem ends with it being epsilon over 8, then the proof continues like normal.

I'm not sure what's happening mathematically here. The PDF on limits has only confused me further, as the notes I have ( which are a copy of the instructor's ) don't mention using a variable such as "M" in them.

Thanks!

2. Originally Posted by Wolvenmoon
Okay, so I did get the problem right the first try. However I am not sure what's going on.

I have the problem: Give an epsilon-delta proof of the limit(x^2-5x-3) = 3 as X approaches -1.

At a point in the proof I come to a point where I have ...|(x-6)(x+1)|<E

My notes and book have some really, really wild stuff going on. Basically "Agree that delta is less than or equal to 1", then it does something really wild. The problem ends with it being epsilon over 8, then the proof continues like normal.

I'm not sure what's happening mathematically here. The PDF on limits has only confused me further, as the notes I have ( which are a copy of the instructor's ) don't mention using a variable such as "M" in them.

Thanks!
You have $\lim_{x \rightarrow -1} x^2-5x-3 =3$

$|(x^2-5x-3)-3| < \epsilon$ If $0<|x+1|<\delta$

That's just using the definition to prove the limit. The first part means that value of your function $f(x)=x^2-5x-3$ will fall between $3- \epsilon$ and $3+ \epsilon$. The second bit just means that x lies in the interval $(-1-\delta , -1+\delta)$. To complete your proof you only need to discover a value of $\delta$ for which this statement holds, and then you must prove that the statement holds for $\delta$.

By the way, $|(x^2-5x-3)-3|= |x^2-5x-6|$ $= |(x-6)(x+1)| < \epsilon$

It's just been factorized, if you were wondering where it came from.