Can You Solve This Epsilon-Delta Proof of a Function's Limit?

[nietzsche]

Homework Statement

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

<br /> \begin{equation*}<br /> f(x) = \left\{<br /> \begin{array}{cc}<br /> 0 &amp; : x \in A\\<br /> x &amp; : x \not \in A<br /> \end{array}<br /> \end{equation*}<br />

Prove that

<br /> \lim_{x \to a} f(x) = a<br />

Homework Equations

The Attempt at a Solution

Proof:

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

which is a continuous function. Thus

<br /> \lim_{x \to a} f(x) = f(a) = a<br />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

<br /> \lim_{x \to a} f(x) = f(a) = a<br />

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.

 

[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 &lt; \epsilon &lt; \dfrac{1}{n(n+1)} then take \delta = \epsilon.

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

 

[nietzsche]

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

 

[HallsofIvy]

Homework Statement

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

<br /> \begin{equation*}<br /> f(x) = \left\{<br /> \begin{array}{cc}<br /> 0 &amp; : x \in A\\<br /> x &amp; : x \not \in A<br /> \end{array}<br /> \end{equation*}<br />

Prove that

<br /> \lim_{x \to a} f(x) = a<br />

Homework Equations

The Attempt at a Solution

Proof:

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

which is a continuous function. Thus

<br /> \lim_{x \to a} f(x) = f(a) = a<br />


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

<br /> \lim_{x \to a} f(x) = f(a) = a<br />

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.

. 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&lt; |x-a|&lt; \delta, then, since |x- a|< delta so |f(x)- a|= |x- a|< \epsilon

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

If a= 0, then, for any \epsilon&gt; 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.

 

[nietzsche]

Thanks very much for the clarification.

 

[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 <br /> <br /> \begin{equation*}<br /> f_{a}(x) = \left\{<br /> \begin{array}{cc}<br /> f(x) &amp; : x \neq a\\<br /> \alpha &amp; : x = a<br /> \end{array}<br /> \end{equation*}<br />

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}?