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Given two numbers $d,h\in\mathbb{R}$ that are governed by this constraint: $$ dh = 1 $$ we know that: $$ h = \frac{1}{d} $$

Therefore $h$, in the limit as $d \rightarrow 0$, is: $$ h = \lim_{d \rightarrow 0} \frac{1}{d} $$

Q1: Is it perfectly fine to say that: $$ h = \infty $$

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  • $\begingroup$ "We do not always define $0\times \infty$ but, when we do, we say it's $0$." $\endgroup$
    – user228113
    Jun 12, 2016 at 8:30
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    $\begingroup$ @G.Sassatelli Thank you. Who are you quoting in your comment, please? It looks to me like a rather weird definition if taken without context, as it can lead to rather serious mistakes in precisely limits, though in some other context, perhaps projective geometry or topology, it could make some sense. $\endgroup$
    – DonAntonio
    Jun 12, 2016 at 8:34
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    $\begingroup$ @Joanpemo A friend of mine. I did it because there is an abused joke about "I don't always do that, but when I do...". The convention I talk about is the one consistently used in measure theory: integral of anything over a negligible set is $0$, the measure of $A\times B$ is $0$ as soon as one of $A,B$ is negligible, et cetera. I agree that this might not be the context. $\endgroup$
    – user228113
    Jun 12, 2016 at 8:44
  • $\begingroup$ @G.Sassatelli Thank you. I see the joke in it. $\endgroup$
    – DonAntonio
    Jun 12, 2016 at 8:55

4 Answers 4

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If you mean $h$ and $d$ to be particular (but unspecified) numbers, then it doesn't make sense to talk about limits at all.

I will henceforth assume you meant for $h$ and $d$ to be real-valued variables (that satisfy $hd=1$)


Your third step is wrong. Something correct would be something like (in no particular order)

$$ \lim_{h \to \infty} h = \lim_{d \to 0} \frac{1}{d} = \lim_{d \to 0} h = \lim_{h \to \infty} \frac{1}{d} $$

where I assume you're working in the projective number line. What we then conclude can be phrased as

  • $h \to \infty$ as $d \to 0$

which can be read as "$h$ approaches $\infty$ as $d$ approahces $0$".

If you mean the extended number line (which is more common for calculus, or at least ideas based on it), you'd have

$$ \lim_{h \to +\infty} h = \lim_{d \to 0^+} \frac{1}{d} = \lim_{h \to +\infty} \frac{1}{d} = \lim_{d \to 0^+} h$$ $$ \lim_{h \to -\infty} h = \lim_{d \to 0^-} \frac{1}{d} = \lim_{h \to -\infty} \frac{1}{d} = \lim_{d \to 0^-} h$$

and a sample conclusion would be

  • $h \to +\infty$ as $d \to 0^+$

or

  • $h \to +\infty$ as $d \to 0$ from the right.

(note that $\lim_{d \to 0} 1/d$ does not exist)


Your mistake might be easier to see in the following flawed argument

  • $x = x$
  • $x = \lim_{x \to 0} x$
  • Therefore, $x = 0$
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  • $\begingroup$ I'm lost. Why did you explain all this if the limit $\lim_{d\rightarrow 0}$ doesn't exist? Doesn't this mean that the only correct answer is to say **limit on right side is $+\infty$, but limit on left side is $-\infty$, and since $+\infty \ne -\infty$, then a limit doesn't exist. Is this your point? $\endgroup$
    – caveman
    Jun 12, 2016 at 9:53
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    $\begingroup$ @caveman: Yes, that's one way to explain why it doesn't exist. $\endgroup$
    – user14972
    Jun 12, 2016 at 10:01
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To say that the limit of $h$ is $\infty$ is to say that $h$ eventually exceeds any real number: for all $r \in \mathbb R$, we have some $\delta > 0$ such that $|d| < \delta$ implies $h > r$.

Since $d$ can approach $0$ from either side, it would be more accurate to say that $|h| \to \infty$.

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No, it is perfectly wrong to say that since you already took $\;h\in\Bbb R\;$ and there is no $\;\infty\;$ element in the real line.

Now, if you took the extended line or some other set that'd include a point $\;\infty\;$ then perhaps you could call $\;h\;$ like that, yet in this particular case even that would be wrong as $\;\lim\limits_{d\to0}\frac1d\;$ doesn't even exist as the one sided limits are different...

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    $\begingroup$ An argument can be valid even if it yields contradictory results -- we call such a thing a proof by contradiction and conclude one of its hypotheses must be false. $\endgroup$
    – user14972
    Jun 12, 2016 at 8:49
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    $\begingroup$ @Hurkyl Thank you. I can't see how an argument can be valid if it leads to contradiction: I think that'd be an oxymoron as "valid", as far as I know, is anything that makes impossible that assumptions are true but the conclusion is false, and this precisely is the basis for the proofs by contradiction or reductio ad absurdum. $\endgroup$
    – DonAntonio
    Jun 12, 2016 at 8:57
  • $\begingroup$ @Hurkyl Exactly: both, but not the argument being valid and results (of it) contradictory. This was my point. $\endgroup$
    – DonAntonio
    Jun 12, 2016 at 9:01
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No. Initially you consider two fixed numbers, $h,d$ that satisfy the relationship $hd=1$

Then you go about considering one as a variable and the other dependant on that variable, thus considering the function $$h(d)d=1\Rightarrow h(d)=\frac1d\\ d\in\Bbb{R} $$

It is not even true that $$\lim_{d\rightarrow 0}h(d)=\infty$$

What does hold is that $$\lim_{d\rightarrow 0^+}h(d)=+\infty\\\lim_{d\rightarrow 0^-}h(d)=-\infty$$

Hence, there is no $d\in\Bbb{R}$ such that $h(d)=\infty$

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