deleted neighborhood of limit L

The question is this true or false. Justify.

Let f:[a,b]->$\displaystyle \Re$ and let c $\displaystyle \in$ (a,b). If lim_{x->c} f(x)=L and L>0, then there is a deleted neighborhood N*^(c) such that f(c)>0 for all x $\displaystyle \in$ N*(c).

If this is true, is the reason because of the following theorem.

Let f: D->$\displaystyle \Re$ and let c be an accumulation point of D. Then lim_{x->c} iff for each neighborhood V of L there exists a deleted neighborhood U of c such that f(U$\displaystyle \cap$D) $\displaystyle \subseteq$ V.

Re: deleted neighborhood of limit L

Right. We only have to choose $\displaystyle V=(0,2L)$ .

Re: deleted neighborhood of limit L

If I understand your terminology correctly, I think that using anything more than the definition of the limit is overkill. If f(x)->L as x->c, and L>0, then choose epislon = L/2. This is a standard type of scenario, both the problem and its solution, and so this standard approach is the best way to do it.

By the definition of the limit, when epsilon = L/2 (note it's positive), there's some positive delta that defines a corresponding deleted open neighborhood U of c

( U = [a,b] intersect (c-delta, c+delta) intersect {c}-complement )

such that for all x in U, abs(f(x)-L) < L/2. Thus for all x in U, -L/2 < f(x)-L, and thus f(x) > L/2. Since L/2 > 0, have shown that there exists a deleted open neighborhood U of c such that for all x in U, f(x) is positive.

Re: deleted neighborhood of limit L

Quote:

Originally Posted by

**johnsomeone** If I understand your terminology correctly, I think that using anything more than the definition of the limit is overkill. If f(x)->L as x->c, and L>0, then choose epislon = L/2. This is a standard type of scenario, both the problem and its solution, and so this standard approach is the best way to do it.

By the definition of the limit, when epsilon = L/2 (note it's positive), there's some positive delta that defines a corresponding deleted open neighborhood U of c

( U = [a,b] intersect (c-delta, c+delta) intersect {c}-complement )

such that for all x in U, abs(f(x)-L) < L/2. Thus for all x in U, -L/2 < f(x)-L, and thus f(x) > L/2. Since L/2 > 0, have shown that there exists a deleted open neighborhood U of c such that for all x in U, f(x) is positive.

Note that $\displaystyle 0 < \left| {x - c} \right| < \delta$ is a deleted neighborhood $\displaystyle (x-\delta,x)\cup(x.x+\delta)$ of $\displaystyle x$.

Re: deleted neighborhood of limit L

Yes. I wrote the deleted nbhd part as "(c-delta, c+delta) intersect {c}-complement" to avoid a more complex statement containing both unions and intersections when full description of the set included the domain [a,b]. That's why I wrote it as "U = [a,b] intersect (c-delta, c+delta) intersect {c}-complement". I thought it might be better than "[a,b] intersect ( (c-delta,c) union (c, c+delta) )", but it's probably not. It's probably more confusing.