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Infinitesimals don't exit

Name: Anonymous 2018-07-09 2:24

In mathematics, they teach us about infinitesimals. But does such a concept even make sense in the physical word?

Name: Anonymous 2018-07-11 3:02

>>16
Isn't -1/infinity a fixed point though?
What do you mean by that?
It's 0.

z=x/infinity approaches 0 as x approaches zero
Harder to visualize when it's written like that.
Time to bust out muh \(\LaTeX\) and BBCode skills!

\(\displaystyle{\lim_{x \to 0} \frac{x}{\infty} = 0}\)

But it doesn't even need to be a limit as x approach zero. It could even be a limit at infinity, just as long as the denominator has a higher power in the dominating x term than that of the numerator.
Example:

\(\displaystyle{\lim_{x \to \infty} \frac{x}{x^{2}} = \lim_{x \to \infty} \frac{1}{x} = \frac{1}{\infty} = 0}\)

For the above, coefficients don't matter as long as the numerator's dominating term's power is less than that of the denominator.

\(\Large Indeterminate \;\; Forms\)

The following are indeterminate forms:

\(\frac{0}{0}\) and \(\frac{\infty}{\infty}\)

However, the above indeterminate forms can be used with L'Hospital's Rule, at least when dealing with derivatives.

\(\displaystyle{\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{\frac{d}{dx}[f(x)]}{\frac{d}{dx}[g(x)]}} = \)\(\displaystyle{ \lim_{x \to c} \frac{\frac{d}{dx}(\frac{d}{dx}[f(x)])}{\frac{d}{dx}(\frac{d}{dx}[g(x)])}} = \; ... \; = \;\)\(\displaystyle{ \lim_{x \to c} \frac{\frac{d^{n}}{dx^{n}}[f(x)]}{\frac{d^{n}}{dx^{n}}[g(x)]}}\)

Not to be confused with the quotient rule, which is different and doesn't apply here.

Other indeterminate forms:
\(0 \times \infty\), \(1^{\infty}\), \(\infty - \infty\), \(0^{0}\), and \(\infty^{0}\).

But for the other indeterminate forms, you're fucked, unless you can somehow factor shit out of them or find a common denominator to rewrite them as determinate forms (or, just get them into \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)to apply L'Hospital's Rule).

And you can keep on applying that until you get a determinate form (as long as \(g^{n}(x) \neq 0\)).

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