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[Spehss _]

Would infinity/zero and zero/infinity be indeterminate forms?

[Bouchart]

infinity/zero is indeterminate.  zero/infinity is zero.

[Spehss _]

zero/infinity is zero.

...right. Can't believe I didn't realize that before posting. Because zero divided by anything except zero is zero. Duh.

New question. So ln(x)=y where e^y=x.

Then ln(0)=y makes it so y is the number where e^y=0. You can't raise anything to a power to get 0 except 0, right? And ln(0)=/=infinity because e^infinity is infinity. What's ln(0) supposed to be? I'm supposed to find the limit of lim x->0(ln(x)/x) using L'hospital's rule. I know how to use the rule, just want to make sure I understand what ln(0) represents.

[TheBiggerFish]

...What is this sorcery?

[Spehss _]

...What is this sorcery?

Indeterminates. Stuff involving zero or infinity.

As I understand it, when solving for lim(f(x)/g(x)) and f(x) approaches 0 or infinity and g(x) approaches 0 or infinity as x approaches a, then the limit is 0/0 or infinity/infinity or infinity/0 or etc, which "may or may not exist" as described by my textbook, so it's called an indeterminate form. And L'Ho(s)pital's rule is a thing that lets you solve for these limits by using derivatives of f and g.

It's, like, y'know, magic and stuff.

Sort of. ln(0) is undefined. Remember that ln(x) = integral from 1 to x of 1/t dt. It's clearly undefined at x=0. ln(0) doesn't represent anything. Now, you could say that lim(x->0+)(ln(x))=-infinity, but the question you've been given clearly doesn't ask for any particular directionality in its approach.

Though I am curious as whether you derived that from some earlier step in a problem or whether that equation was explicitly given, because you'll quickly find that using L'Hopital's rule is going to leave you in a rather nasty situation.

Also, it's L'Hopital, not L'Hospital.

Whoops. Left out the plus. Question was lim(x->0+)(ln(x)), and the book the textbook gives is negative infinity. Looking at how I solved it in an earlier homework assignment and I just went with ln0=0 to get 0/0, then solved with derivatives and L'ho's rule from there to get (1/x)1=1/x which substituted is 1/0=infinity.

And my book says L'hospital. ¯\_(ツ)_/¯

[Bouchart]

lim x->0 ln(x) is -infinity.  Easiest to see on a graph, but remember that for x<1, ln(x)<0.  Take ln(1/2)=y -> ln(1) - ln(2)= y -> -ln(2)=y

lim x->0 ln(x)/x = lim x->0 (1/x)*(1/x)=(1/x^2)= infinity

[frostshotgg]

infinity/zero is indeterminate.

Infinity/0 is undefined, not indefinite.

[Bouchart]

No, it's called an indeterminate form, which means it has a value but it takes a bit more effort to find out what it is.

[Reelya]

Actually, Bouchart, indeterminate value is the correct terminilogy, but they definitely don't have a value that "takes a bit more effort to find". It's easy to find the value, except you can generate an infinite number of contradictory values:

The most common example of an indeterminate form occurs as the ratio of two functions, in which both of these functions tend to zero in the limit, and is referred to as "the indeterminate form 0/0". As x approaches 0, the ratios x/x3, x/x, and x2/x go to ∞, 1, and 0 respectively. In each case, if the limits of the numerator and denominator are substituted, the resulting expression is 0/0, which is undefined. So, in a manner of speaking, 0/0 can take on the values 0, 1 or ∞, and it is possible to construct similar examples for which the limit is any particular value.

So, 0/0 can be proven to have any value: between, and including negative and positive infinity. To get an answer of "0/0 = 2" for example, you take the limit as x approaches zero of y=2x/x, and similarly for any real value you care to name. So 0/0 is simultaneously all possible values.

[frostshotgg]

No, it's called an indeterminate form, which means it has a value but it takes a bit more effort to find out what it is.

Did you actually read your own link?

The expression 1/0 is not commonly regarded as an indeterminate form because there is not an infinite range of values that f/g could approach.

By extension, any non-0 value in the numerator is not an indeterminate form. It's merely undefined. In that case, it's a really, really large number, or a really, really large negative number.

[Furtuka]

Could anyone explain to me how Asymtotic Analysis works in the context of discrete math? I've been looking at readings on it over and over and just have no idea how the mathematical part of it is suppose to work with the choosing the N and d and c and all that, and how Big Theta, O, and Omega are actually suppose to be found via proofs.

[MagmaMcFry]

Maybe it helps if I describe it: Let's say you have two sequences a(n) and b(n) of positive real numbers. We write a(n) = O(b(n)) (the equals sign is misleading) if a(n)/b(n) is (eventually) bounded from above; a(n) = o(b(n)) if a(n)/b(n) converges to zero; a(n) = theta(b(n)) if a(n)/b(n) is bounded from above and below by two positive numbers (so a(n) = O(b(n)) and b(n) = O(a(n))).

To prove that a(n) = O(b(n)), you just need to take the sequence (a(n)/b(n)) and prove that it is bounded from above by finding a number that is larger than all values in the sequence.

Note that if your sequences contain some zeros or negative numbers in the beginning, you may simply cut off the beginning of your sequences until a(n)/b(n) is well-defined.

[MaximumZero]

I understand the words in that series...

[Baffler]

I got this question wrong on my physics exam. Can anyone explain how to solve this? You might need more physical constants. I'm copying this from memory, we didn't actually get this thing back to keep.

Say you have a 200g copper cup that contains 150mL of water, both at 20°C. An ice cube weighing 30 grams at 0°C is dropped in. What is the final temperature of the system? The specific heat of copper is 0.385 J/g°C, the specific heat of liquid water is 4.186 J/g°C, the specific heat of ice is  2.03 J/g°C. The latent heat of fusion of water is 334 J/g.

[TheBiggerFish]

Hold on just a second.  The ice cube can't be 20 degrees Celsius, it would be water...?


But I think (I haven't done this since last year) that you would want to sum up the total energy of the system, subtract out the energy needed for the phase change from ice to water if you hit it, and then...Yeah, i think that's your answer?


Again, this is merely an educated guess.