Calculus, Analytical demonstration

araveugnitsug

1. The problem statement, all variables and given/known data
Demonstrate Analytically:
If [itex]\lim_{x\to a}f(x)=L[/itex] and L<0, there exists a δ>0 so that x[itex]\in[/itex]Dom(f), 0<|x-a|<δ [itex]\Rightarrow[/itex] f(x)<0

2. The attempt at a solution
I start identifying what I want to demonstrate and what are my assumptions.
x[itex]\in[/itex]Dom(f), 0<|x-a|<δ [itex]\Rightarrow[/itex] f(x)<0
I'm aware I have to reach f(x)<0 from the fact that L<0 and using the definition of a limit, the x belonging to the domain is there so that it also covers the cases were if not stated, one could get a limit that does not exist due.

So I have:
0<|x-a|<δ and 0<|f(x)-L|<ε
and I suspect the way is not in the delta part but working with 0<|f(x)-L|<ε, but because epsilon is in relation to delta it never banish, meaning I would have to work with the other part.

0<|f(x)-L|
Because L<0 is part of the hypothesis
L<|f(x)-L|+L
L-|f(x)-L|<L
Triangle Inequality
L-|f(x)|-|L|<L
Because L<0 is part of the hypothesis
L-|f(x)|-(-L)<L
L-|f(x)|+L<L
L-|f(x)|<0
-|f(x)|<-L
L<-f(x)<-L
-L>f(x)>L
And I don't know where else to go to get the f(x)<0 form, I simply have no way to get rid of the L in the process without being left with an absolute value on f(x).

HallsofIvy

Using the defintion of limit, take [itex]\epsilon[/itex] to be any number less than L.

araveugnitsug

It works wonderfully and makes it two simple steps. Thanks.

Though I'm inclined to ask; is this something that can always be done, correlating epsilon with the actual limit?