The point of the thread was to confuse some people who don’t like math.

Still, perpetual motion cannot happen, it violates the law of...what’s its name...thermodynamics? Law of Entropy, that’s it.

Didn’t fool me. Inifinity has some interesting concepts, though.

0/0 = ∞
∞/∞ = ∞ etc...

Still, perpetual motion cannot happen, it violates the law of...what’s its name...thermodynamics? Law of Entropy, that’s it.

0/0 and ∞/∞ are indeterminate forms; that is, they may approach any finite number or +/-∞ depending on how f(x)/g(x) behaves. L’Hospital’s Rule can come in handy for these types of situations. It basically says that if we want to calculate

lim{x->a} f(x)/g(x),

and f(x) and g(x) both approach 0 or both approach +/-∞ in that limit, then

lim{x->a} f(x)/g(x) = lim{x->a} f’(x)/g’(x).

However there may be cases where even the repeated use of l’Hospital’s Rule won’t help you.

Examples:

Type 0/0:
lim{x->1} (ln x)/(x - 1) = lim {x->1} (1/x)/1 = lim {x->1} 1/x = 1.

Type ∞/∞:
lim{x->∞} e^x/x^2 = lim{x->∞} e^x/(2x) = lim{x->∞} e^x/2 = ∞.

In the second example you apply l’Hospital’s Rule twice. Alternatively you can take a shortcut by noting e^x grows faster than x^2.

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Now how about a situation where

lim{x->a} f(x) = 0 and lim{x->a} g(x) = ∞,

and you want to find lim{x->a} f(x)g(x)? We can write

fg = f/(1/g) or fg = g/(1/f)

and use l’Hospital’s Rule again.

Example:

Type 0*∞:
lim{x->0+} x ln x = lim{x->0+} (ln x)/(1/x) = lim{x->0+} (1/x)/(-1/x^2) = lim{x->0+} -x = 0.

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So in a nutshell, it all depends.

As for 0^∞, until I see a convincing argument/proof otherwise, I believe it does equal 0. However just something to chew on: 1^∞ is actually an indeterminate form.