How can I define an infinitely small positive value?

The short answer is that you can't. But you can define any positive value which is arbitrarily small, in other words as small as you want it to be.

This means that just like how there is no "largest number" (since for any number $N$, we know that $N + 1$ is larger), there is no smallest positive number, since for any real number $N \in \mathbb{R}$,

$$\frac{1}{N} > \frac{1}{N+1} $$


The thing is $\infty$ doesn't refer a number that is "infinitely big." There is no such thing, nor does it mean a number that is arbitrarily large -- such a number would still be finite -- nor does it refer to any actual values.

It is somewhat incorrect to say things like $1+1+1+.... = \infty$ or $\frac 10 = \infty$ or $\frac 1\infty = 0$. Those are all incorrect statements.

More technically the correct way to state those would be as limits:

$\lim_{n\to \infty} \underbrace{1+1+1+.... +1}_{n\text{ times}} = \infty$, or $\lim_{x\to 0^+} \frac 1x = 0$ and $\lim_{x\to \infty} \frac 1x = 0$.

The phrase "$\lim_{x\to c} somethingexpressedintermsof(x) =k$" means.... well, informally, it means, "as $x$ gets closer and closer to $c$ then $somethingexpressedintermsof(x)$ gets closer and closer to $k$".

For example $\lim_{x\to 5} x^2 = 25$.

Now if instead of getting "closer and closer" to something specific, we want to express the idea that the values get "larger and larger" and that, if you imagine any number as larger as you want this value will at some point get that large and even larger than that -- that's when we use the symbol $\infty$.

BUT THIS IS IMPORTANT! The symbol doesn't represent any actual value. It only represents the idea that a series of values can get larger and for any large value you can imagine the series of values will somewhere get as large as that.

So $\lim_{x\to \infty} \frac 1x = 0$ means if we take $x = 1,000$ then $\frac 1x$ will be close to $0$ and if we take a larger value like $x= 10^{1,000}$ or $x=10^{googol}$ or $x$ getting "larger and larger" we get $\frac 1x$ gets larger and larger.

Alternatively $\infty$ can be in the "detonation spot". $\lim_{x\to 5}\frac 1{x-5} = \infty$ means this: if we take $x$ really really close to $5,$ then $\frac 1{x-5}$ will be really really large. And we can make $\frac 1{x-5}$ as large as we like by taking $x$ as close to $5$ as we like.

.....

Okay, so if $\infty$ is the symbol for getting things really really large what is the symbol for getting things really really small. Well, ... that means "getting close to $0$". We had that all the time.

$\lim_{x\to 0} \frac {x^2 + 4x}{x}=4$ means "if we take $x$ to be as arbitrarily small as we like". And that's pretty much the concept you want. There's no such thing as $\infty$ or a number that is "infinitely large" and there is no such thing as a number that is "infinitely small". And if a number getting infinitely large "tends toward infinity", the number getting infinitely small will "tend toward zero".

And that's pretty much it.

... HOWEVER...

We want to be able to do math on these small values -- these values that are arbitrarily close to zero but not quite at zero.

When we do that we frequently say: "Let $\epsilon > 0$ be arbitrarily small" Or "Let $\delta$ be an arbitrarily small positive number" or "let $h > 0$". In these cases though, it's important to realize that although we are trying to figure out what happens when these variables get "infinitely small", at no time or value are these numbers anything strange or different. These variables represent actual normal numbers. Just very small ones.

For example, consider this problem: if $f(x) = x^2 + 5,$ what is $\lim_{x\to 5} \frac {f(x) -f(5)}{x-5}$?

Let $|h| > 0$ be arbitrarily small. Then if $x = 5 + h$ we have

$\frac {f(5+h) - f(5)}{(5+h) - 5} = \frac {(25 + 10h + h^2) - 25}{h}= \frac {10h + h^2}{h} = 10 + h$.

So $\lim_{x\to 5} \frac {f(x) -f(5)}{x-5}$

$= \lim_{h\to 0} \frac {f(5+h) - f(5)}h $

$ = \lim_{h\to 0}10 + h = 10 + 0 = 10$.

Notice that although $h$ is our "infinitely small" number, it is an actual number.

Also note that if $h$ does actually equal $0$ we do not get $\frac {f(5+h) - f(5)}h = 10$. We get $\frac {f(5+h) - f(5)}h=\frac 00$ is undefined garbage. This only works if $h$ is "arbitrarily small" and we get $\frac {f(5+h)-f(5)}h = 10 + h,$ and as $h$ is "arbitrarily small" this $\to 10$ as $h\to 0$.

Tags:

Infinity

Intuition

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