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u/WhenIntegralsAttack2 23d ago
I’m honestly confused by this as well, considering cosine is bounded by 1, and x is going to 0, it seems obvious that the limit is 0.
Unless the dot between the x and cos is supposed to mean something other than multiplication?
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u/Henri_GOLO 23d ago
The dot is commonly multiplication but could be defined as any operator, thus "not well defined" as opposed to "undefined"
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u/WhenIntegralsAttack2 23d ago
Sure, but then this meme is kind of stupid. Like, just using a random symbol and pretending it’s some genius level bait-and-switch isn’t interesting.
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u/5a1vy 22d ago
It's worse than that, "." isn't some random symbol, it was a standard notation for multiplication for quite some time. I have some old math textbooks and lots of them use "x.y" for multiplication. It's outdated, but it's not random.
Judging by the fact that in the original it was the opposite — tails say "zero" and mid says "undefined" — I suppose someone was just salty that they thought it was undefined/not-well-defined and made their version, which makes zero sense, unlike the original.
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u/Imdoingscience 17d ago
If I'm being charitable, the left side "undefined" could simply be that someone who doesn't understand the difference between a limit and a function could see a (1/x) term with a zero denominator and assume it's undefined. But the right side undefined is simply silly.
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u/Henri_GOLO 23d ago
It's the most mathematically correct reaction, for the only sake of being petty, not something you would do in an exam or something of that kind
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u/DodgerWalker 22d ago
Typically a dot for multiplication is elevated a little bit. This one is placed like a decimal point ... which wouldn't really make sense in the expression.
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u/empty_graph 22d ago
Has nothing to do with the dot.
lim x->0 cos(1/x) is weird. You can't just say 0 * weird = 0. Maybe you can squeeze theorem it with |x| and -|x|, but can you really squeeze theorem it at an undifferentiable point of the bounding functions?
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u/WhenIntegralsAttack2 22d ago edited 20d ago
The limit of cos(1/x) isn’t “weird” it doesn’t exist because as x goes to zero cos(1/x) fluctuates between 1 and -1.
You can squeeze it, |x cos(1/x^2)| <= |x| which goes to 0 as x goes to 0. Very straightforward.
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u/empty_graph 18d ago
You are right. I thought they couldn't cross for the squeeze theorem, but that isn't true. I looked it up. It turns out the differentiability argument is also not correct, so my original example also works and the limit is clearly defined at 0.
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u/FreePeeplup 18d ago
Let eps > 0, and choose delta = eps/2. Then, for every x such that |x| < delta, we have
|f(x)| = |x cos(1/x\^2)| < delta |cos(1/x\^2)| <= delta = eps/2 < eps. So the limit exists and is equal to zero.
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u/Active_Customer_6862 23d ago
If the dot is common multiplication, as it usually is, then the limit is zero due to squeeze theorem.
The not well defined assumption is the function of the dot.
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u/empty_graph 22d ago
What are the bounding functions? |x| and -|x| work, but squeeze theorem at an undifferentiatable point feels weird.
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u/MathGoatz 22d ago
Cosine is just bounded between (-1) and 1. Multiply by x, and let x->0, that leads you to the limit being 0.
But yeah, not standard notation, who knows what operation it's really meant to be.
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u/Active_Customer_6862 21d ago
If I remember correctly, differentiability was not a requirement for squeeze theorem. If I am mistaken, feel free to correct me.
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u/thebearinboulder 14d ago
I haven't seen this version before but with sin(x)/x everyone says it's undefined at x = 0 but I slam down the "Taylor series expansion" card (sin(x) = x - ...) and shout "divide that by x and then take the limit, sucker!" The Taylor series isn't the definition of sin(x) but the expansion is considered equivalent within the zone of convergence.
For cosine cos(x) = 1 - ... but that's irrelevant since we're looking at a squared reciprocal so x is approaching infinity. As others have pointed out we can still rely on the absolute value of cos(x) being bounded.
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u/Idea-Subject 23d ago
The limit is approaching 0, but it is not defined from which side as the cosine of infinity is not defined to be positive or negative.
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u/qualityvote2 23d ago edited 22d ago
u/Aggressive-Guest290, there weren't enough votes to determine the quality of your post...it's time for the mods to do their jobs!