# Another epsilon-delta proof

1. Oct 17, 2009

### nietzsche

1. The problem statement, all variables and given/known data

Similar to the problem I just posted, here's another one:

Suppose $$A$$ is the set of all numbers that can be written as $$\dfrac{1}{n}$$, where $$n \in \mathbb{N}$$.

Define a function $$f$$ such that

$$\begin{equation*} f(x) = \left\{ \begin{array}{cc} 0 & : x \in A\\ x & : x \not \in A \end{array} \end{equation*}$$

Prove that

$$\lim_{x \to a} f(x) = a$$

2. Relevant equations

3. The attempt at a solution

Proof:

For $$x \in (-\infty, 0)\cup(1,\infty) \Rightarrow f(x) = x$$

which is a continuous function. Thus

$$\lim_{x \to a} f(x) = f(a) = a$$

Now consider the interval $$(0, 1]$$. Suppose $$a$$ is in this interval. If $$a \not \in A$$, then $$\delta$$ can be chosen sufficiently small so that the open interval $$(a-\delta, a+\delta)$$ does not contain any $$x \in A$$. In this subinterval, $$f(x) = x$$ and

$$\lim_{x \to a} f(x) = f(a) = a$$

Now suppose $$a \in A$$. Then $$a$$ can be written as $$a = \dfrac{1}{n}$$.
Choose $$\delta = |\dfrac{1}{n}-\dfrac{1}{n+1}| = |\dfrac{1}{n(n+1)}|$$.
Thus $$\delta$$ is the minimum distance to the next $$x \in A$$.

Once I get here, I'm stuck... I'm trying to show that if delta is the minimum distance to the next x in A, then for all $$x \not = a$$ in that subinterval, $$f(x) = x$$. But I'm confused about what to do next. I'm not sure what this implies with regards to epsilon.

I'm also not sure what to do for x = 0.

Last edited: Oct 18, 2009
2. Oct 18, 2009

### nietzsche

I've had a think about it and I think this is what it means...

If $$\epsilon \geq \dfrac{1}{n(n+1)}$$ then take $$\delta = \dfrac {1}{n(n+1)}$$.

If $$0 < \epsilon < \dfrac{1}{n(n+1)}$$ then take $$\delta = \epsilon$$.

Ugh, the way I've chosen to do it seems complicated...

3. Oct 18, 2009

### nietzsche

can anyone confirm that what i'm on the right track here?

4. Oct 18, 2009

### HallsofIvy

. There is nothing much else to be done. For that delta, as you say, $f(x)= x$ so |f(x)- a|= |x- a|. Now, if we call that "delta" $\delta_1$, we can take $\delta$ to be the smaller of $\delta_1$ and $\epsilon$. If $0< |x-a|< \delta$, then, since |x- a|< delta so |f(x)- a|= |x- a|< $\epsilon$

If a= 0, then, for any $\epsilon> 0$, there exist x in A less than $\epsilon$ so that f(x)= 0. But now |f(x)- a|= |f(x)| so that's not a problem.

5. Oct 18, 2009

### nietzsche

Thanks very much for the clarification.

6. Oct 18, 2009

### aPhilosopher

Both of these problems are trivial in light of one lemma. The set up for the lemma is this: Let $$f$$ be a function on a set $$S$$ and suppose that $$\lim_{x \to a} f(x) = b$$

Now let $$\begin{equation*} f_{a}(x) = \left\{ \begin{array}{cc} f(x) & : x \neq a\\ \alpha & : x = a \end{array} \end{equation*}$$

Where $$\alpha$$ is arbitrary (in fact, we don't even need to define $$f_{a}$$ at $$a$$). What can you say about $$\lim_{x \to a}f_{a}$$?