# Show that f(x) is unbounded

## Homework Statement

Show that the function
$$f(x) =\\ \begin{cases} \frac{1}{x} &\quad 0 < x \leq 1\\ 0 &\quad x = 0 \end{cases}$$

is unbounded.

## Homework Equations

If f is bounded, |f(x)| <= M for all x in f's domain.

## The Attempt at a Solution

I tried arguing by contradiction: suppose there is such an M. Then |f(x)| = f(x) <= M. But if f(x) < 1/M, f(x) > M. But I get stuck because that means this particular choice of bound does not work. Instead, choose N > M. But then f(x) < 1/N makes f(x) > N. There might be a bound, and I'm having trouble proving that there is not a bound.

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Mark44
Mentor

## Homework Statement

Show that the function
$$f(x) =\\ \begin{cases} \frac{1}{x} &\quad 0 < x \leq 1\\ 0 &\quad x = 0 \end{cases}$$

is unbounded.

## Homework Equations

If f is bounded, |f(x)| <= M for all x in f's domain.

## The Attempt at a Solution

I tried arguing by contradiction: suppose there is such an M. Then |f(x)| = f(x) <= M. But if f(x) < 1/M, f(x) > M.
If x < 1/M, then f(x) > M.
You can do this directly, without resorting to a proof by contradiction.
bigplanet401 said:
But I get stuck because that means this particular choice of bound does not work. Instead, choose N > M. But then f(x) < 1/N makes f(x) > N. There might be a bound, and I'm having trouble proving that there is not a bound.

If x < 1/M, then f(x) > M.
You can do this directly, without resorting to a proof by contradiction.
But then can't you choose N > M and still be bounded? I'm guessing that I have to show that we can always take x small enough so that there's no M that will always satisfy the condition f(x) <= M for x in (0, 1]. But I don't know how to do that.

Mark44
Mentor
But then can't you choose N > M and still be bounded?
I think you might be confusing this problem with one in which $\lim_{x \to \infty} f(x) = \infty$.
bigplanet401 said:
I'm guessing that I have to show that we can always take x small enough so that there's no M that will always satisfy the condition f(x) <= M for x in (0, 1]. But I don't know how to do that.
Yes. Let some "large number" M be given. Then if 0 < x < 1/M, then f(x) > M. That's all you need to say.

LCKurtz
@bigplanet401: The statement that $f(x)$ is bounded means that there is a number $M$ such that for all $x$, $|f(x)|\le M$. I think it would help your thought process if you wrote a careful denial of that. The statement that $f(x)$ is unbounded means..., then prove that.