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Let $f(x) = \frac{1}{x}$ be a function to and from the reals. Is $f$ bounded?

I want to prove using the definition. Let $M > 0$ be given. Then there exists a $\delta > 0$ such that:

If $0<|x-0| < \delta$ implies $|f(x)|=|\frac{1}{x}|=\frac{1}{|x|}>\frac{1}{\delta} = M$ whenever $\delta = \frac{1}{M}$

So is it correct to pick $\delta = \frac{1}{M}$ and I have proved that as $x$ tends to $0$ which, $f(x)$ tends to positive infinity, so it is unbounded.

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  • $\begingroup$ What is the definition you're using for "bounded"? $\endgroup$
    – Sambo
    Feb 21 at 18:29
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    $\begingroup$ Small correction: the natural domain and range of the reciprocal function is $\mathbb{R}^* = \mathbb{R} \setminus \{0\}$. $\endgroup$ Feb 21 at 18:32
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    $\begingroup$ @adisnjo The converse of that definition is "for all $M>0$, there exists some $x_M$ (in the domain) such that $|f(x_M)| \geq M$. So, once you let $M>0$ be given, you should tell us which value $x_M$ we should pick for this inequality to hold. $\endgroup$
    – Sambo
    Feb 21 at 18:35
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    $\begingroup$ @adisnjo That's correct. In that case, your proof could be clarified by adding something like the following line: "To prove $f$ is unbounded, it suffices to show that $\lim_{x \rightarrow 0} |f(x)| = \infty$, so this is what we'll do." You could also adjust it to say: "picking $\delta = 1/M > 0$, we find that $0 < |x-0| < \delta$ implies...". This makes it clear what your choice of $\delta$ is from the start. $\endgroup$
    – Sambo
    Feb 21 at 19:07
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    $\begingroup$ However, in my personal preference, a direct proof would be better since it would be more succint. $\endgroup$
    – Sambo
    Feb 21 at 19:08

2 Answers 2

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The formulas that you've written actually prove something stronger, namely that $$ \lim_{x \to 0}\; \biggl\lvert \frac1x \biggr\rvert = \infty. $$

To establish that $f(x) = \frac1x$ is unbounded, we have to show that it is either unbounded from above or from below. The following applies to the upper bound, but everything works mutatis mutandis to show that for this function there's no lower bound either.

For each potential upper bound $M$, we have to demonstrate that there's some $x$ in the domain of $f$ such that $f(x) > M$. Your formulas show that it suffices to choose any $x \in \smash{\bigl(0, \frac1M \bigr)}$, but such a universal quantifier isn't strictly necessary here. In other words, we just need an example point $x$ rather than all possible $x$ in a neighborhood of $0$.

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  • $\begingroup$ Thanks, but my proof does imply that $f$ is unbounded. Thanks for your help. $\endgroup$
    – adisnjo
    Feb 21 at 18:58
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If $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ was bounded there would exist a constant $M >0$ such that $$ |f(x)| < M, \quad \forall x\ne 0. $$

However, this generates a contradiction since $|f(1/M)| > M$.

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