Oh, all right.. No delta because x is extending without bound towards negative infinity, but there is an epsilon because the function approaches a finite value of L, correct?
I understand that.
But, I still don't know how I'd go about constructing a proof.
Say you want to show,
You need to show that for any
We can choose an
Such that if,
is in the domain of the function.
Implies,
Since both side are positive we can take the reciprocal of both sides, but then we have to flip the inequality, obtaining an equivalent statement,
That means, .
We also note that,
is in the domain of .
So this tell us given any we need to choose
wow, major headache.. I understand all your examples
Thing is. The only thing that's killing me on mine is that there is no specific function... I'm just given f(x)
In your example, you wrote
as
That isn't too bad to do because f(x) is defined as a specific function,
I don't know how I'd write in the form x < something, when I don't have any specific function, so I only have an f(x), not an x.
*sigh*