Cap'n Refsmmat

No. Inertia doesn't push or pull on you. It just makes it harder for other things to push or pull you around.

Pretty much 99.9% of the time. There might be a situation where kinetic is greater, but I can't think of any.

It doesn't "go" anywhere. It's just that it no longer acts. Those little imperfections aren't meshing -- they're bouncing against each other now that it's moving. You could say it's the same thing happening (imperfections grinding against each other), just that when the surfaces start moving they bounce, making the amount of friction go down. Hence the change from one coefficient to another.

Depends on what units you're using.

This was one of those semi-artificial "healthy" butter things, so I'm guessing it did have more than the usual amount of water. So what would be a good way to prevent this? I'm guessing the first option is to take the melted butter off the heat soon after it has completely melted to keep it from continuing to heat up. Any other ways to stop it?

So if we look at a point on the "end" of the domain of the function -- such as 0 in our example -- we cannot take the limit because we do not have two "sides" to approach from? My understanding was that you just took the limit from the "side" you have and called it good. I will have to dig out my textbook tomorrow and see what else I'm mixing up with limits.

That seems contradictory. If [math]\lim_{x\to a} (f(x) \times g(x)) = \lim_{x\to a}( f(x)) \times \lim_{x\to a} (g(x))[/math], it's pretty much implied that you can take the limits individually -- why else would you have split the limit in two? I'd like to see the explanation for this. I guess there are lots about limits I have yet to know...

Why is that? From what I understand of limits, the following is true: [math]\lim_{x\to a} (f(x) \times g(x)) = \lim_{x\to a}( f(x)) \times \lim_{x\to a} (g(x))[/math] and you can evaluate each part safely.

No, that's not [math]\frac{\infty}{\infty}[/math] because [math]\lim_{x \to 0} \frac{1}{\sin x}[/math] doesn't exist. It turns out the natural log method would have worked if we were cleverer: [math]\ln y = \lim_{x\to 0}\sin x \ln x[/math] [math]\ln y = \lim_{x\to 0} \frac{\sin x}{(\ln x)^{-1}}[/math] apply L'Hopital's Rule, since the above limit is indeterminate: [math]\ln y = \lim_{x\to 0} \frac{\cos x}{-(\ln x)^{-2} \times x^{-1}}[/math] Simplificate: [math]\ln y = -1 \times \lim_{x\to 0} \cos x \times \lim_{x\to 0} (\ln x)^2 \times x[/math] [math]\ln y = -1 \times \lim_{x\to 0} (\ln x)^2 \times x[/math] [math]\ln y = -1 \times \lim_{x\to 0} \frac{(\ln x)^2}{x^{-1}}[/math] L'Hopital again: [math]\ln y = -1 \times \lim_{x\to 0} \frac{2\ln x \times x^{-1}}{-1x^{-2}}[/math] [math]\ln y = -1 \times \lim_{x \to 0} \frac{2\ln x}{-x^{-1}}[/math] ...and again: [math]\ln y = -1 \times \lim_{x \to 0} \frac{2x^{-1}}{x^{-2}}[/math] [math]\ln y = -1 \times \lim_{x \to 0} 2x[/math] [math]\ln y = \lim_{x \to 0} -2x[/math] [math]\ln y = 0[/math] [math]y = 1[/math] If anyone sees an error, please tell. I'd like to think I've conquered this one (with a bit of help). And a question: I find it hard to believe [math]\lim_{x \to 0} \frac{\ln x}{x^{-1}} = \frac{\infty}{\infty}[/math]. The bottom doesn't have a definite limit (I think) as you approach zero, so can you safely call it [math]\infty[/math] anyway to make the limit indeterminate?

You think something as impure as liquid butter can superheat? I mean, it's got lumps in it and stuff. From what I recall with water superheating, you usually need a clean mug and pure water to make it work particularly well.

We just had a rather interesting experience. While melting butter on the stove on low heat, the liquid butter essentially exploded, sending hot butter raining all over the kitchen. This doesn't make much sense to me. What exactly would cause something on low heat to gather enough energy to splatter the entire kitchen? (One of our cats immediately decided the floor tasted rather good. That was funny.)

I've noticed that a lot on Wikipedia, especially in math. I'll go to check something and it'll take me a few minutes just to figure out what the heck the article is saying. I'm guessing there are a few too many upper-level math students writing articles.

The trouble is, we haven't learned about series expansions yet. So is there an alternative method or was this question really just shown to us too soon? What we've been learning, and the section of the textbook this problem was from, is L'Hopital's Rule. Our approach was to try to get the problem into the form [math]\frac{0}{0}[/math] or [math]\frac{\infty}{\infty}[/math] and work from there, but there's no clear way to do that with this problem. If it could be done, it'd make a lot more sense.

So essentially what's happening is we decide (as in your example) that [math]\lim_{x \to 1} \cos x = 1[/math] and thus we can substitute in 1 in the original limit? Or is there some limit technique I don't yet know?

Yeah, [math]0^0[/math] is indeterminate. I guess you can take the limit though, as ajb points out. It's deceptively simple. Thanks, everyone.

Today's calculus warmup question has stumped everyone in the class, including our teacher. Here's the question: [math]\lim_{x\to 0} x^{\sin x}[/math] We can't think of any safe way to evaluate the limit. We tried rewriting the problem to this: [math]y = \lim_{x\to 0} x^{\sin x}[/math] and then changing to this: [math]\ln y = \lim_{x\to 0}\sin x \ln x[/math] but in the end it didn't help. Any ideas? [hide]The answer turns out to be 1, but we have no idea to get it.[/hide]

It seems to be working for me. Is this the link you were trying?

Chuck Norris.

I'm guessing that's partly due to the fact that we've made learning painful.

They don't. As far as I know, brain dead people are considered dead, because when you're brain-dead not even the involuntary things (like breathing) are occurring.

Heterosexuals are allowed to marry the person they love the majority of the time, whereas homosexuals aren't. Do we really have to get as technical as discussing incest? The point's the same either way.

So it's an incentive designed to not make students want it? I don't really see the point. If it's an effective incentive -- meaning people will want to get the money -- then you instantly get the "ooh, money, must get good grades" effect. You can't have the incentive without it. Learning has enough benefits by itself. It should be able to serve as its own reward. Indeed. But I still think there's a better system possible beyond straight-up grades.

An incentive makes a student focus on achieving something. What you're describing is simultaneously an incentive and not. Why should effort even be part of this? Find ways of accurately judging how well a student has learned and can use material -- not just through regurgitation, but through thinking. Effort is included by default: it takes effort to understand and be able to use new knowledge. Although we don't really know, because nobody can audit their budget to see what can be cut.

Bad idea. In the state of Texas, the top 10% of every graduating class is guaranteed admission into any Texas public college. The valedictorian gets his or her first year free. What does that do? It means competition is incredibly fierce and people get into fancy class-juggling where they adjust their schedules to give them the best possible GPAs. You wouldn't believe the angst people get over their grades. I'm of the view that grades are a really dumb idea in general. Rather than repeat my arguments, I'll just shamelessly plug my blog.

Define community service. I think that will be a major problem with this scheme. People will find crap activities that they can call "community service" and just get around the system.